# EXPONENTS_ LOGS_ GEOMETRIC SEQUENCES _ SERIES

Document Sample

```					        CHAPTER 2 & 6 – EXPONENTS, LOGS, GEOMETRIC SEQUENCES & SERIES

DAYS 1-2

Review Exponent Laws                               Examples of Each Law

1. xm  xn = xm + n                                         a3  a 7  a10

xm                                         a3                   1
2.     n
 x mn                                  7
 a 37  a 4  4 (avoid negative exponents!)
x                                          a                   a

3. a. (xm)n = xmn
b. (xmyn)c = xmcync                        a  3 4
 a34  a12
c
 xm  x mc
c.  n   nc
y    y

1                                     1
4. a.       xa                                a 3 
xa                                    a3
1
b.       a
 xa
x
a
x
a
 y  ya
c.                  or a
 y              x   x

5. a. x0 = 1
b. x = x1

6. a. (x)2 = (x)(x)
b. x2 = (1)(x)(x)

                                      a
m                                        3
m                                      3
7. x n  n x m or          n
x                a 4  4 a3       4

Page 1                                 M30P LP CH 2 & 6 Logs Exp
Examples: Simplify. (Positive Exponents!!)

27x3                   (2 xy 3 ) 2 (3 x 3 y )3                 3y 7
1. (3xy2)3                          ANS:                       2.                                     ANS:
y6                        (6 x 7 y ) 2                         x3

3
 6 x 5 y                                   y3
3.        2 
ANS:                       4. 70  22 + 21 + 3x0 + (8x)0           ANS: 1.5
 2 xy                                     27 x12

2                                                       5
                            4                                                               1
5. 31  27           3
ANS:                       6. 81        4
ANS:
9                                                               243

82 x  3 2 x  5
7.                                    ANS: 23 x20
16 x  4

Assignment
Simplify (positive exponents only)
(3 xy )3 (4 x 3 y 5 )                   3y14
1. (2x3y)4                         ANS: 16x12y4                  2.                                     ANS:
(6 x 4 y 3 ) 2                       x2

3
3    2                    x6                          10 x 6 y 7                             125x18
3. (2x y)                         ANS:                          4.                                   ANS:
4 y2                         2y
9
                              y6

1
5. (8 x12 y 6 ) 3                 ANS: 2x4y2                   6. 80 + (7x)0  6x0 + (2)4  22           ANS: 8

3                5
1
7. 16 4  (8) 3                   ANS: 24                      8. ( 3 8) 5                           ANS:
32

9 x 3 27 2 x 7
9. 25x  7  87  x ANS: 22x + 14                                10.                                    ANS: 34 x35
81x  2

3
 20 x 2 y 6                          1                        (  xyz 2 )3 (2 x 3 z ) 4               2z 7
11.                              ANS:                         12.                                    ANS:
 5 xy 
4
64x3 y 6                       ( 2 x 5 y 3 z ) 3                   y6

Extra:
pg. 333 # 1 – 12
pg. 71 # 1 – 36

Page 2                                   M30P LP CH 2 & 6 Logs Exp
DAYS 1-2

A. Solving Exponential Equations
B. Solving Equations with Exponents Other Than 1

When the bases are the same, (or can be manipulated to be the same) and the exponent is in the variable,
we can solve by comparing exponents.       Remember if ax = ay then x = y.

Examples:
43
1. 32x  7 = 3x + 2                    ANS: 9               2. 93x  4 = 27x  17                       ANS: 
3
Bases are the same, exponents equal                          Get bases to be the same
2x  7  x  2                                               93 x4  27 x17
3 
2 3 x 4
  33 
x 17
x9
3       3
6 x 8        3 x 51

6 x  8  3 x  51
3x  43
43
x
3

43 x  2                                   10                          1
3.          1                         ANS:                 4. 5x 4                                   ANS: –7
8x2                                        3                         125

43 x  2  8 x  2
1
2 
2 3 x2
  23                                         5x4 
x2
53
6 x  4  3x  6                                             5 x  4  5 3
3x  10                                                      x  4  3
10                                                      x  7
x
3
2
5. x3 = 8                              ANS: 2               6. x 5  4 ans : 32

2
What number to the power of 3                                What number to the power of
5
equals 8?                                                    equals 4?

1                                                              5
Or, x  8     3
Or, x  4       2

Page 3                                   M30P LP CH 2 & 6 Logs Exp
5
 25  2
3                                                            2
                                     26
7. 2( x  1)       2
 250                    ANS:                  8. 3x  11  14
5
ANS:   or 200.47
25                                                         3 

3                                                                2

( x  1)           2
 125        divide by 2                             3 x 5  25
2
2
x  1  125 3                                                                           25
x5 
1                                                                    3
x 1                                                                                        5
25                                                            25  2
26                                                                   x     200.47
x                                                                               3 
25

9. 8x+ 3 = 1                                    ANS: –3

8x+ 3 = 1
8x+ 3 = 80                              Anything to the power of zero is 1
x+ 3= 0
x= - 3

Assignment
pg. 89 # 1, 3, 6, 9, 10, 11, 12
plus:

1. Solve for x:                      (* nearest hundredth)

4                                                1
1
a.            x  16
3
ANS: 8                b. 2 x    2
 10              ANS:
25

2                                       5
c. (2 x  3) 3  9 ANS: 15                               *d. 7 x 6  1  57                  ANS: 12.13

3                                           2
                                1
*e. (7 x  2)  5 ANS: 2.375
f.   3x       3
 7  19          ANS:     or 0.125
8

Page 4                                        M30P LP CH 2 & 6 Logs Exp
DAYS 1-2

Review

1. Simplify
a  a 2 x          3 y

=
a  a 
2 x    3 y


 a  a 
2x        3y


a 2 x 3 y
 a3 y
a a                                              a a              a a
x y          x y                            x y    x y          x y       x y
a2 x
m a  b mb  c                                    m a b mb  c m a  2b  c
2. Simplify                a  c 
=  a  c    a  c  m 2 a  2 b
m                                                m            m

2 x 5 2
 4x       5 x
2
3. Solve for x

2 
x 5 2
 4x            5 x
2

22 x 10   22 
x 2 5 x

22 x 10  22 x         10 x
2

2 x  10  2 x 2  10 x
0  2 x 2  12 x  10
0  x2  6x  5
0  ( x  5)( x  1)
x  5,1
2
 27      3

 64 
                                                                                                                                   9
4. Solve          1
In calculator                          (27 / 64) ^ (2 / 3) /(125 ^ (1/ 3)) 
3
80
125

Exponential Function – this is a short lesson. Feel free to start the day with some review type
questions.
 y = ax, a > 0

2 Basic Situations

1. Start low and the shoot up.                                                      2. Start high and shoot down.

Page 5                                                      M30P LP CH 2 & 6 Logs Exp
Examples:

I. Graph and find:

1.   x–int
2.   y–int
3.   horizontal asymptote (what is non–permissible value for y?) What y value is never reached?
4.   domain (always x  R ) An exponent can be positive or negative.
5.   range

a. y = 2x                                          b. y = 5x

1.   none                                          1.   none
2.   (0, 1)                                        2.   (0, 1)
3.   y=0                                           3.   y=0
4.    xR                                          4.    xR
5.   y>0                                           5.   y>0

c. y = 2x  3                                      d. y = 2x + 3        Need  x  3 in brackets!

1.   (1.58, 0)                                     1.   none
2.   (0, 2)                                       2.   (0, 8)
3.   y = 3                                        3.   y=0
4.    xR                                          4.    xR
5.   y > 3                                        5.   y<0

x
1
e. y = 5x + 1 or y =   + 1                      f. y = 3x  1  27
5

1. none                                            1. (4, 0)
80
2. (0, 2)                                          2. (0,  )
3
3. y = 1                                           3. y =  27
4. x  R                                           4. x  R
5. y > 1                                           5. y > 27

g. y = 3  2x         Note:  is the same as 

1.   none
2.   (0, 3)
3.   y=0
4.    xR
5.   y>0

Page 6                          M30P LP CH 2 & 6 Logs Exp
Assignment
Graph and find:
1. x–int
2. y–int
3. horizontal asymptote
4. domain
5. range

1.                         2.                3.                  4.                  5.
A. y = 2x + 5
B. y = 3x  2  9
C. y = 5x + 1
x
1
D. y     27
3
E. y = 2x  2  2
F. y = 5  2x

y = ax        yY bx y = cx y = dx
1
Plus:                                     =                             We know the values of a, b, c, and d to be 2, 3,         ,
10                                                                                         3
9                                  and 6 (but not in this order). Which is which?
8
7
6
5
4
3
2
1
X

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1            1 2 3 4 5 6 7 8 9 10
1
-2                                                           ANS: a =     , b = 6, c = 3, d = 2
-3                                                                      3
-4
Plus pg. 76 # 2 – 6         -5
-6
-7
-8
-9
1.
-10                         2.                        3.                  4.                 5.
x
A. y = 2 + 5                     none                 (0, 6)                     y=5                 xR                y>5
x2
B. y = 3         9                                          8                  y = 9
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
(4, 0)                                                               xR               y > 9
(0, 8 )
9
C. y = 5 + 1x
(0, 0)                (0, 0)                     y=1                 xR                y<1
1
x
(3, 0)              (0, 26)                   y = 27               xR               y > 27
D. y     27
3
E. y = 2x  2  2               (3, 0)                       3                  y = 2               xR                y > –2
(0, 1 )
4
F. y = 5  2    x
none                (0, 5)                     y=0                 xR                y<0

Page 7                           M30P LP CH 2 & 6 Logs Exp
DAY 3      Short Lesson (Start Day 4 as Well)

Determining Equation of Exponential Function
(Algebraically & Using the Regression Function on Calculator)

Regression Function
 curve of best fit

We will find the exponential function equation in the form y = abx. Most problems can be solved using
this form and the calculator uses this form only.

Example:

1. Exponential function of the form y = abx contains the points (1, 10) and (3, 250). Find a & b.

250  ab3
10  ab
25  b 2
5  b (b  0) 10  a(5), 2  a

y = 2(5)x

* This can be done using the graphing calculator:    Hit Stat then hit Edit on your calculator

1. clear lists L1 and L2

2. enter x values in L1 and y values in L2

3. you can graph the points using stat plot (you may have to change the window)

4. go to stat, then calc, then exponential regression (you do not need to enter L1 and L2 as they are
defaulted)

5. if r2 and r do not appear, you have to turn diagnostic on by hitting 2nd 0 (catalog) and selecting
Diagnostic On (enter must be hit twice)

* The r value gives a coefficient correlation which shows how well the equation represents the data (1 is
a perfect correlation)

Page 8                         M30P LP CH 2 & 6 Logs Exp
1.
The population of an ant colony was studied over time.

Day      Population of
Colony
0            98
1           268
2           681
3          1411
4          2622

a.   What is the exponential regression equation that best fits the data?
ANS: y  111.92  2.28x

b.   What would you expect the ant population to be on day 15?
ANS: y  111.92  2.2815  26 177 635

2.   Use an exponential regression equation for the data below.

Wolf                 What is the exponential regression equation?
Year
Population
0             500
1             548
2             800                                       ANS: y  458.79  1.34 x
3            1 168
How many wolves would you expect in year 5?

ANS: y  458.79  1.345  1982

Page 9                              M30P LP CH 2 & 6 Logs Exp
Assignment

1. Determine the exponential function given points (1, 6) and (4, 48) on the function. Do both ways.
ANS: y = 3  2x

2. Tuition Fees. Use the data below.

a. Determine the exponential function for each using regression function.
ANS: Educ. y = 873.80  1.12x
Eng. y = 1077.34  1.20x

b. Estimate tuition fees in 2005 for Education.                    ANS: \$4783

Education             Engineering
1991 (1)           \$990                  \$1300
1992 (2)          \$1094                  \$1539
1993 (3)          \$1223                  \$1865
1994 (4)          \$1358                  \$2251
1995 (5)          \$1574                  \$2681

3.
Joe invested \$5000 in an account and tracked the growth.

Year (L1)    0     1     2     3     4     5
Value (L2) \$5000 \$5350 \$5725 \$6125 \$6554 \$7013

a. Write the exponential regression equation that best models this data.
ANS: y  5000  1.07 x

b. How much would he have after 10 years?
ANS: y  5000  1.0710  \$9835.76

Page 10                          M30P LP CH 2 & 6 Logs Exp
DAY 4

Logarithms        (Critical Day = Must Know Cold)

Log Form                                               Exponential Form

y  log b x                                             by  x

*   b > 0, b  1 (will be explained when we get to functions)

Common Logs – base of 10 (log 5; base 10 is understood)
We write log 5 instead of writing log10 5 . When no base is written, 10 is assumed.

Natural Logs – base e (ln 5; e 2.71828 )
We write ln 5 instead of writing ln e 5 . When no base is written, e is assumed.

Examples:

1. Change to exponential form.

a. logmp = r                 ANS: mr = p           b. log232 = 5             ANS: 25 = 32

c. logm10 + 2 = x            ANS: mx  2 = 10      d. 7 = log3(4x  5) ANS: 37 = 4x  5

Rewrite as
log m 10  x  2

2. Change to log form.

a. 34 = 81                   ANS: log 3 81  4     b. mc = d                 ANS: log m d  c

c. (x + m)3 = a + b          ANS: logx + m(a + b) = 3

3. Evaluating logarithms.
5
a. log381 = x                ANS: 4                b. log432                 ANS:
2

3x = 81                                            4 x  32
3x = 34   x=4                                      2 
2 x
 25
5
2x  5          x
2

Page 11                             M30P LP CH 2 & 6 Logs Exp
1
c. log 2                           ANS: –6                 d. log 5 25                   ANS: 4
64

x

2x 
1                                                5  25
64                                              x4
1
2x 
26
2 x  26              x  6

5
e. log3 9 3                        ANS:                    f. log1111                    ANS: 1
2

11x  11
3x  9 3
x 1
1
3x  32  32
5
5
3 3
x         2
x
2

g. log51                           ANS: 0                  h. log m m n                  ANS: n

5x  1                                                     mx  mn
x0                anything to power 0                     xn
equals 1

5                                                              7
10                                                  7
i.   log        a7                 ANS:                    j.   log    4   m10           ANS:
a3                               21                                                  8
m5

x           5
a3  a 7
4x       7
3x            5
m  m10
5
a a
2             7
4x 7
3x 5                                                           
                                                        5 10
2 7                                                       40 x  35
21x  10
35 7
10                                                   x      
x                                                             40 8
21

Page 12                                M30P LP CH 2 & 6 Logs Exp
Assignment for Day 4
pg. 98 – 99 # 1, 4, 7, 10, 13, 16, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32

plus:
11                                                10
33                                                  50
1. log 5 a 2                ANS:                  2. log 5 a a 7                ANS:
a3                         10                                                   7

10
x
5
a a7
5x        11
a a23                                                 1x        10
a a 5          7
5 x 11
                                             x 10
3     2                                             
10 x  33                                         5 7
7 x  50
33
x                                                          50
10                                             x
7

Worksheet B 1- 5 is good for review as well.

Page 13                                M30P LP CH 2 & 6 Logs Exp
DAY 5      (Shorter Lesson)                          Can Start Laws of Logs Day 6

Solving Basic Log Equations

Determine Exact:
3                    2
1. log168 = x              ANS:                   2.      log x 9                  ANS: 27
4                    3
2

16  8
x                                             x3  9
3

2 4 x
 23                                    x  92                   If you get 364.5, you typed
2 4 x  23                                        x  27
4x  3
3                                                                    9 ^ 3/ 2 not 9 ^ (3/ 2)
x
4

3              1                                       4
3. log 25 x              ANS:                   4. log   8
x                   ANS: 4
2             125                                      3

4
3
83  x
25   2
x
4

x  0.008 
1                                     1 3      4

8   x  8  4
2         6
125
 
Don’t type 25 ^  3 / 2

Solve for x  0.01 :

5
1. log7x = 1.73            ANS: 28.97             2. log x 294                     ANS: 30.27
3

5
71.73
x                                      x  294
3

3
x = 28.97                                          x  294  30.27     5

Page 14                                    M30P LP CH 2 & 6 Logs Exp
2                                                          2        
3.    log (2 x 7) 29         ANS: 74.58         4. 3  log  2x 3  5  ANS: 11 264.30
3                                                                   
2
2                                    103  2 x 3  5
 2x  7   3    29                                          2
3
1000  2 x 3  5
2 x  7  29    2
2

2 x  7  156.17                                1005  2 x 3
2 x  149.17                                                 2
502.5  x 3
x  74.58                                                    3
x  502.5 2  11264.30

Page 15                            M30P LP CH 2 & 6 Logs Exp
Assignment for day 5
pg. 99 #33 – 40

plus: Solve for x  0.01

3
1. log11x = 0.43                     ANS:      2.80   2.   log x 4                   ANS: 25.40
7

3

11 0.43
x                                              x 4
7

x  2.80                                                      7
4  x  25.40
3

5
3. log (2 x3) 16                   ANS:      4.14   4.   log 3.4 (7 x  11)  1.9   ANS: 3.03
3

5
 2 x  3 3  16
3.41.9  7 x  11
3
2 x  3  16 5
10.23  7 x  11
2 x  3  5.28
21.23  7 x
2 x  8.28
x  3.03
x  4.14
2

5. log 6 (3 x       5
 3)  1.3     ANS: 0.11

2
61.3  3 x 5  3
2
10.27  3 x 5  3
2
7.27  3 x 5
2
7.27
x5
3
2
2.42  x 5
5
2.42 2  x
x  0.11

Page 16               M30P LP CH 2 & 6 Logs Exp
DAY 6

Laws of Logarithms                                       Formulas 1–4 are on formula sheet.

M      
1. loga  M  N   loga M  loga N                   2. log a          log a M  log a N
N      

log a c
3. log a M n  n log a M                              4. log b c 
log a b

5. log x x  1          * not on formula sheet

6. log x x k  k        * not on formula sheet        (This is based on #3 and #6)

7. M logM x  x      ( M log M x  y,   log M y  log M x,  y  x) * not on formula sheet

Examples:

1. Express as a single logarithm and evaluate whenever possible.

a. log 250 + log 4                    ANS: 3         b. log354  log32                    ANS: 3

log  250  4                                               log 3  54  2 
log1000  log10 1000                                         log 3 27
10  1000
x
3x  27
x3                                                          x3

 ab 
c. logx5x3 + logx3x7  logx15x        ANS: 9         d. log a + log b  log c             ANS: log  
 c 
 5 x3  3x 7 
log x                                                      No work needed. Rules of logs.
 15 x 
15 x10
log x
15 x
log x x 9
x a  x9
a9

 ab 
e. log a  log c + log b                         ANS: log  
 c 

No work needed. Rules of logs.

Page 17                              M30P LP CH 2 & 6 Logs Exp
1                                                          p 
f.     log p  2 log m  3log r                       ANS: log  2 3 
2                                                         m r 
     
1
log p  log m 2  log r 3
2

log p  log m 2  log r 3
 p 
log  2 3 
m r 
     

g. 2log 3x + 3log 2x2                                 ANS: log (72x8)

log(3 x) 2  log(2 x 2 )3
log(9 x 2 )  log(8 x 6 )
log  72 x8 

2. Expand the following.

mx 2
a. log 3                                              ANS: log3 m  2 log3 x  log3 p
p

3
x2 y5                                      2         5
b. log p                                              ANS: log p x  log p y  3log p m
m3                                         3         3

2   5
3
x2 y5         x3 y3 2         5
log p           3
 log p    3
 log p x  log p y  3log p m
m              m    3         3

3. Given log3 x  5 and log3 y  8 , evaluate:
9x2
a. log 3 xy           ANS: 13                         b. log 3         ANS: 4
y
log 3 x  log 3 y                                    log3  9 x 2   log3 y
5  8  13                                           log 3 (3 x) 2  log 3 y
2 log 3 (3 x)  log 3 y
2  log 3 3  log 3 x   log 3 y
2 1  5  8  2  6  8  4

Page 18                           M30P LP CH 2 & 6 Logs Exp
y
c. log 3            ANS: 3
3
1
log 3 y 2  log 3 3
1
log 3 y  log 3 3
2
1
 8 1  4 1  3
2

4. Given log54 = x and log53 = y, express in terms of x and/or y.

a. log 5 12         ANS: x + y                     b. log 5 45              ANS: 2y + 1

log 5 4  log 5 3                                 log 5 3  log 5  3  log 5 5
x y                                              y  y 1  2 y 1

4                                                                         2
c. log 5            ANS: x  2y                    d. log5 3 32             ANS:       y
9                                                                         3

2
log 5 4  log 5 9                                 log 5 3   3

log 5 4   log 5 3  log 5 3                    2
log 5 3
x   y  y                                      3
x  2y                                            2
y
3

1
e. log52            ANS:      x
2

log 5 4
 1
log 5  4 2 
 
1
log 5 4
2
1
x
2

Page 19                          M30P LP CH 2 & 6 Logs Exp
5. Simplify.

a. 3log3 7                          ANS: 7                 b. mlogm y                                    ANS: y

Rule #5                                                   Rule #5

c. 4log2 5                          ANS: 25                d. x3log x 2 y                                ANS: 8y3

 22 
log x  2 y 
3
log 2 5
x
 
log x 8 y 3
22log 2 5                                                 x
 
log 2 52                                               8 y3
2
2 log 2 25
 25

e. 7log7 3log7 y                   ANS: 3y

7 log7 (3 y )
7 log7 (3 y )
 3y

Assignment
pg. 106 # 1, 3, 7, 11, 15, 18, 19, 22, 25, 26, 29, 30, 32, 33, 53 a – c
plus:

# I of Worksheet A

1. m2logm 5                           ANS: 25              2. 9log3 2 x                  ANS: 4x2

m    log m (52 )                                            3       
2 log 3 2 x

log 3  2 x 
2
m log m 25                                                32log3 2 x  3
 25                                                                
log 3 4 x 2
3                        4x2

3. 5log5 x  2log5 y                  ANS: xy2

log5 x  log5 y 2 
5
 
log5 xy 2
5
 xy 2

Page 20                                              M30P LP CH 2 & 6 Logs Exp
DAY 7

Solving Complex Log Equations Using Properties

*   Either Log = Number (or algebraic expression) or Log = Log

Examples:

1. log3 x  log3 (2 x  10)  log 3 3                     ANS: 6

log 3 x  log 3  6 x  30 
x  6 x  30
30  5 x
x6

2. 1  log3 5x  6  log3  x  2                      ANS: 6

1  log 3  5 x  6   log 3  x  2 
log 3 3  log 3  5 x  6   log 3  x  2 
5x  6                                                          5x  6 
1  log 3         
3                                                                  x2 
x2
3x  6  5 x  6                                              5x  6
or       31 
12  2 x                                                       x2
3x  6  5 x  6
x6
12  2 x
x6

13
3. log2 3x 1  log2  x 1  6                        ANS:        ,5
3
log 2  3x  1  log 2  x  1  6 log 2 2
log 2  3x  1  log 2  x  1  log 2  26 
log 2  3x  1  log 2  x  1  log 2 64
 3x  1 x  1  64
3x 2  2 x  1  64
b  b 2  4ac
3x  2 x  65  0
2
use quadratic formula            x
2a
13
x         ,5                  can ' t have negative answer in logs
3

Different Technique below

Page 21                                M30P LP CH 2 & 6 Logs Exp
log 2  3 x  1  log 2  x  1  6
log 2  3x  1 x  1   6
                  
26   3 x  1 x  1
64  3 x 2  2 x  1
b  b 2  4ac
3x 2  2 x  65  0              use quadratic formula            x
2a
13
x          ,5                  can ' t have negative answer in logs
3

4. 3log 2 x  4 log 2 x  21                                 ANS: 8

log 2  x3   log 2  x 4   21log 2 x
log 2  x   log 2  x
3           4
  log  2 
2
21
7 log 2 x = 21
or
x 7  2097152                                               log 2 x = 3                  divide by 7
3
1                                        2 = x= 8
x  2097152 7  8

5. 3  logx  2x  3  logx (x  6)

3log x x  log x  2 x  3  log x  x  6                 3  log x  2 x  3  log x  x  6 
log x  x3   log x  2 x  3 x  6  
                                      3  log x  2 x  3 x  6  
                    
x 3  2 x 2  9 x  18                             or       x3  2 x 2  9 x  18
x 3  2 x 2  9 x  18  0                                  x3  2 x 2  9 x  18  0
x  3 , 2,3                                                x  3 , 2,3

 log3 x         5log3 x  6  0
2
6.                                          (use substitution) ANS: 9 and 27

Let y  log3 x
y2  5 y  6  0
 y  3 y  2   0
log 3 x  3  0                  log 3 x  2  0
log 3 x  3                      log 3 x  2
3 x
3
32  x
27  x                           9x

Assignment
pg. 113 – 114 # 21, 23, 25, 40 a, c, e, f, h

Could also use Worksheet A (assign all)

Page 22                             M30P LP CH 2 & 6 Logs Exp
DAY 8

Solving Exponential Equations Using Logs

Method I: Take log of each side using any base (we always use base 10 because it is on the
calculator)

Method II: Change from exponential form to log form. You can use this method when only one
base has a variable for an exponent.
i.e. only one base has a variable exponent. Questions 1, 4, 5

Examples:

1. 2x = 19         Method I log 2x = log 19                    Method II 2x = 19
xlog 2 = log 19                             log219 = x
log19                                    log19
x=                                                  =x
log 2                                    log 2
x = 4.25                                     4.25 = x

4 log 7  2 log 5                                         log 5  2 log 7
2. 53x  2 = 7x + 4        ANS:                                3. 5  3x = 7x  2     ANS:
3log 5  log 7                                            log 7  log 3
or 3.82                                                  or 6.49

log 53 x  2  log 7 x  4                                   log 5  log 3x  log 7 x  2
 3x  2  log 5   x  4  log 7                           log 5  x log 3   x  2  log 7
3x log 5  2 log 5  x log 7  4 log 7                       log 5  x log 3  x log 7  2 log 7
3x log 5  x log 7  4 log 7  2 log 5                       log 5  2 log 7  x log 7  x log 3
x  3log 5  log 7   4 log 7  2 log 5                     log 5  2 log 7  x  log 7  log 3
4 log 7  2 log 5                                            log 5  2 log 7
x                      3.82                                x                    6.49
3log 5  log 7                                               log 7  log 3

5x
2 log 3 29
4. 3 2  29                             ANS: x =              or 1.23
5

5x
log 3 2  log 29
5x
log 3  log 29
2
5 x log 29

2     log 3
2 log 29 2 log 3 29
x                      1.23
5log 3       5

Page 23                              M30P LP CH 2 & 6 Logs Exp
5
5
5. 13  7   2x
ANS: x =                 or 1.90
2 log 7 13

5
log13  log 7 2 x
5
log13       log 7
2x
log13 5

log 7 2 x
2 x log13  5log 7
5log 7 5log13 7
x                    1.90
2 log13       2

Assignment
pg. 113 # 1, 2, 4, 9, 11, 13, 17, 20
plus: questions from worksheet A or B

Page 24   M30P LP CH 2 & 6 Logs Exp
DAY 9               Can do questions a, b and c in 30 minutes. Great quiz day.

Graphing Log Functions

    Consider the equation y  log b x where b > 0 and b  1
It could also be written as b y  x
    It is the inverse of b x  y , reflected over line y = x

Examples:

1. Find domain, range, x & y–intercepts, equation of asymptote and sketch for:

a. y = log2x      (2y = x)                            b. y = log2(x + 4) + 2

D: x > 0                                              D: x > 4
R: y  R                                              R: y  R
x–int: (1, 0)                                         x–int: (–3.75, 0)
y–int: none                                           y–int: (0, 4)
asymptote: x = 0                                      asymptote: x = 4

Graph in calc as                                           Graph in calc as
y1  log( x) / log(2)                                         y1  log( x  4) / log(2)  2

Domain: although the graph appears to start at (0, –2), it actually    Domain: although the graph appears to start at (–4, –1), it
shoots down the y axis and allows y to get incredibly small.            actually shoots down x = –4 and allows y to get incredibly
Thus D: x > 0.                                                          small. Thus D: x > –4.
Range: The graph shoots down for every y and thus R: y  R .           Range: The graph shoots down for every y and thus R: y  R .
x – intercept: solve for y = 0                                         x – intercept: solve for y = 0
y  log 2  x  4   2
y  log 2 x                                                         0  log 2  x  4   2
0  log 2 x                                                          2  log 2  x  4 
20  x  1                                                                                  1
22  x  4            x 4  3.75
4
Asymptote: since x can not be zero, the vertical                      y–intercept: solve for x = 0
asymptote is given by x = 0                                             y  log2 (0  4)  2
y  log2 4  2  3
Asymptote: since x can not be negative 4, the vertical
asymptote is given by x = –4

Page 25                                   M30P LP CH 2 & 6 Logs Exp
c. y   log3  x  9                                d. y  log4 8  x  1

D: x> 9                                              D: x < 8
R: y  R                                              R. y  R
x–int: (8, 0)                                        x–int: (4, 0)
y–int: (0, 2)                                        y–int: (0.5, 0)
asymptote: x = 9                                     asymptote: x = 8

Graph in calc as                                                           Graph in calc as
y1   log( x  9) / log(3)                                        y1  log(8  x) / log(4)  1

Domain: although the graph appears to start at (–9, 1), it actually    Domain: although the graph appears to start at (8, –2), it
shoots up x = –9 and allows y to get incredibly big. Thus               actually shoots down x = 8 and allows y to get incredibly
D: x > –9.                                                              small. Thus D: x < 8.
Range: The graph shoots down for every y and thus R: y  R .           Range: The graph shoots down for every y and thus R: y  R .
x – intercept: solve for y = 0                                         x – intercept: solve for y = 0
y   log 3  x  9 
y  log 4  8  x   1
0   log 3  x  9 
0  log 4  8  x   1
0  log 3  x  9 
1  log 4  8  x 
30  x  9
41  8  x                x  84  4
1 x9                x  8
y–intercept: solve for x = 0                                          y–intercept: solve for x = 0
y   log3  0  9                                                            y  log 4  8  0   1
y   log3  9   2                                                           y  log 4  8  1  0.5
Asymptote: since x can not be –9, the vertical                        Asymptote: since x can not be 8, the vertical
asymptote is given by x = –9                                      asymptote is given by x = 8

Page 26                                     M30P LP CH 2 & 6 Logs Exp
Day 9 Assignment

Domain     Range       x–int      y–int     asymptote
y = log3(x  2)
y = log5x  1
y = log2(x + 8)  1
y = log3(9  x)
also known as
y = log3((x  9)

Domain    Range       x–int       y–int    asymptote
y = log3(x  2)       x>2        yR      (3, 0)      none         x=2
y = log5x  1         x>0        yR      (5, 0)      none         x=0
y = log2(x + 8)  1   x > 8     yR      (6, 0)     (0, 2)       x = 8
y = log3(9  x)
also known as         x<9        yR      (8, 0)      (0, 2)       x=9
y = log3((x  9)

Page 27                       M30P LP CH 2 & 6 Logs Exp
DAY 10

Application of Logarithms

A. Log Functions

1. Richter Scale                                      a. The Richter scale reading of earthquake A is 6.8
and of earthquake B is 7.9. How many times as
I
R = log10I or (R = log10      , use I o  1)      intense is B compared to A?      ANS: 12.59
Io

R = Richter scale reading                             6.8  log I
I = actual intensity                                  106.8  I  6309573.445

7.9  log I
107.9  I  79432823.47

107.9
6.8
 101.1  12.59
10
or
79432823.47
 12.59
6309573.445

b. Earthquake A has a Richter scale rating of 6.8. Earthquake C was 50 times as intense. What was
earthquakes C Richter scale rating?

6.8  log I              106.8  50  IC  315478672.2
106.8  I A
R  log10 I
R  log10 106.8        or      R  log10 315478672.2
R  8.5                          R  8.5

Page 28                             M30P LP CH 2 & 6 Logs Exp
2. Sound                                           a. The decibel reading of sound A is 79.8. Sound B
is 250 times as intense. What is its decibel
dB = 10log L

dB = decibel reading                                    79.8  10log L
L = actual loudness                                     7.98  log L
107.98  L  95499258.6
3. The sound of the final explosion was heard
over 4500km away and covered 1/13th of the
Earth's surface.                                            107.98  250  2.37 1010
5. The final death toll from pyroclastic flows,

dB  10 log 107.98  250 
volcanic bombs, and tsunamis was calculated
to be a devastating 36,417.
dB  103.78

KRAKATOA VOLCANO ERUPTION-1883 A.D., CRACKED ONE FOOT THICK CONCRETE
AT 300 MILES, CREATED A 3000 FOOT TIDAL WAVE, SOUND PRESSURE CAUSED BAROMETERS TO
FLUCTUATE WILDLY AT 100 MILES
INDICATING LEVELS OF AT LEAST 170-190 DB (P) AT THIS DISTANCE OF 100 MILES
EVEN WHEN SHOUTING IN SOMEONES EAR, COULD NOT BE HEARD AT 100 MILES
CAUSED FOG TO APPEAR AND DISAPPEAR INSTANTLY AT HUNDREDS OF MILES
ROCKS WERE THROWN TO A HIEGHT OF 34 MILES. DUST AND DEBRIS FELL CONTINOUSLY FOR 10
DAYS AFTER BLAST. PRODUCED VERY COLORFUL SUNSETS FOR ONE YEAR, EJECTED 4 CUBIC MILES OF
THE EARTH. CREATED ANTI-NODE OF NEGATIVE PRESSURE AT THE EXACT OPPOSITE SIDE OF THE
EARTH.SOUND COVERED 1 / 10 OF THE WORLDS SURFACE, SHOCK (SOUND) WAVES “ECHOED” AROUND THE
EARTH 36 TIMES AND LASTED FOR ABOUT A MONTH!

3. Welding Glasses                                 a. What shade number would be required so that
approximately 25% of the light would pass
through the glasses?            ANS: 6
20 log T
Sn =             2
3
20log 0.25
Sn = shade number                                   Sn                   2
3
T = fraction of light transmitted through                              Sn  6
glass, if 20%, then T = 0.2
b. If you have shade number 9, what percent of
of the light is transmitted through the glasses?
ANS: 8.9%

20 log T
9             2
3
20 log T
7
3
21  20 log T
21
 log T
20
21
10   20
 T  0.089  8.9%

Page 29                                M30P LP CH 2 & 6 Logs Exp
B. Using Logs to Solve Exponential Functions

1. Exponential Growth

t
A(t )  A0 (2) d

A(t) = resulting number
A0 = original number
t = total time
d = doubling time

Example:

a. Bacteria double every 4 hours. How long before you have 3000 bacteria if you start with 20
bacteria?

t
3000  20(2) 4
t
150  2 4
t
log150      log 2
4
4 log150  t log 2
4 log150
t
log 2
t = 28.92

2. Exponential Decay

t
 1 h
A(t )  A0  
2

A(t) = resulting mass
A0 = original mass
t = total time
h = half life

Page 30                         M30P LP CH 2 & 6 Logs Exp
Example:

a. 60g of radioactive material decays to 20g in 40 years. Find its half life.

40
1h
20  60  
2
40
1 1h

3 2
1 40       1
log        log
3 h        2
1          1
h log  40 log
3           2
1
40 log
h          2
1
log
3
h  25.24

Assignment
pg. 99 # 46 a, d (do not use trial and error), 50 a
pg. 107 # 48 a, b
pg. 114 # 43, 44 a, b
plus:

20 log T
1. Sn =           2
3
a. Find the shade number if 3% of light passes through. (nearest whole number) ANS: 12

20log 0.03
Sn                    2  12
3

b. Find the percent of light that passes through if the shade number is 15. ( 0.01)
ANS: 1.12%

20 log T
15            2
3
20 log T
13 
3
39  20 log T
39
 log T
20
39
10 20  T  0.012  1.12%

Page 31                  M30P LP CH 2 & 6 Logs Exp
DAY 11

Geometric Sequence

Notation
t5 = 5th term
t1 = a = 1st term
tn = general term
S5 = sum of 1st 5 terms
S1 = t1 = a
Sn = sum of 1st n terms

Geometric Sequence
 ratio of consecutive terms is a constant (r = common ratio)
 each term multiplied by a common ratio to get the next term
o 5, 10, 20, 40… each term is multiplied by 2.
 General form
tn = arn  1

Develop the Formula:
2,         6,               18,                        54, ...
2,         2  3,           2  3  3,                 2333
t1         t2               t3                         t4

therefore, tn  ar n 1     tn  2(3) n 1

Examples:

1. 4, 8, 16, ..., Find:

a. t9                   ANS: 1024                       b. tn                ANS: tn = 2n + 1

t9  4(2)91  1024                                        tn  4(2) n 1
tn   22  2n 1
tn  2n 1

c. n for 8 388 608      ANS: 22

8388608  2n 1
log 8388608  log 2n 1
8388608  2n 1
log 8388608  (n  1) log 2
log 2 8388608  n  1
or      log 8388608
23  n  1                                                     n 1
log 2
n  22
23  n  1
n  22

Page 32                            M30P LP CH 2 & 6 Logs Exp
5 2
2.  , ,... is a geometric sequence. Find:
8 3

16
a. r (exact)                 ANS:                   b. t25 (0.0001) ANS: 2.9415
15

2
251
t2     16                                               5  16 
r  3                                             t25                      2.9415
t1 5 15                                                 8  15 
8

3. 0.6, 0.72, ... is a geometric sequence. Find:

6
a. r ANS:           or 1.2                           b. smallest value of n so that the term is  57
5
ANS: 25.977, so n = 26

t2 0.72 6
r            1.2                                tn  ar n 1
t1   0.6 5
57  0.6(1.2)n 1
95  1.2n 1
log1.2 95  n  1
24.977  n  1
25.977  n  26

4. t3 = 12, t7 = 192. Find a & r. (substitution or elimination)                  ANS: 3 and 2 or 3 and 2

12  ar 31       192  ar 7 1
12  ar 2         192  ar 6
12                192
a                 a
r2                 r6
t3 12
t2       6
12 192                                              r 2
 6
r2   r                                             t    6
a 2      3
12r  192r 2
6
r 2
r 6 192

r 2 12
r 4  16
r  2

Page 33                         M30P LP CH 2 & 6 Logs Exp
Geometric Means
 term(s) between 2 terms of a geometric sequence so that together they form a geometric sequence

Examples:

1. Find 2 geometric means between 5 & 135.                                     ANS: 15, 45

5, ______, _______, -135

135  5r 41
135  5r 3
5 3  15           153  45
27  r 3
r  3

2. Find the missing terms: ____, 4, ____, 8                               ANS: 2 2, 4 2 or  2 2, 4 2

tn  ar n 1
4   2  4 2, 4 2
8  4r 31
4   4    4 2 4 2
2  r2                                     4 2      ,        ,      2 2, 2 2
2  2    2    2
r 2

Example:

1. Find x so that x  1, x + 1, and 2x  1 form a geometric sequence.            ANS: 0, 5

t 2 t3

t1 t2
x  1 2x 1

x 1 x 1
 2 x  1 x  1   x  1 x  1
2 x 2  3x  1  x 2  2 x  1
x2  5x  0
x( x  5)  0
x  0,5

Page 34                       M30P LP CH 2 & 6 Logs Exp
Assignment for day 11
pg. 300 – 301 # 1, 3, 7, 8, 12, 17, 20, 23 a, c, 30 a
plus:

1. 3, 3.6, ... is a geometric sequence. Find:

a. t52 (0.01) ANS: 32 761.58                   b. smallest value of n so that the term is at least 200
ANS: n = 25

t    3.6                                         200  3(1.2) n 1
r 2       1.2
t1   3                                          200
 1.2n 1
tn  ar n 1                                         3
 200 
t52  3(1.2)51                                     log1.2         n 1
 3 
t52  32761.58
n  24.03  25

1
2. x  2, x + 3, 3x  1 form a geometric sequence. Find x.                           ANS:  , 7
2

t 2 t3

t1 t2
x  3 3x  1

x2 x3
 3x  1 x  2    x  3 x  3
3x 2  7 x  2  x 2  6 x  9
2 x 2  13x  7  0
1
x  ,7
2

Page 35                          M30P LP CH 2 & 6 Logs Exp
DAY 12       (Less than a day)

Application of Geometric Sequence

Compound Interest
 A  P 1  i 
n

i = interest rate per compounding period
n = # of compounding periods

Example:

A              P                r            compounding          time
period
1.                 ?            \$5000          10%/a             quarterly         3 years
2.              \$9000           \$3000          9.3%/a             monthly             ?
3.                5x              x               ?            semi–annually      18 years

A              P                r            compounding          time
period
1.         (\$6724.44)
2.                                                                              (143 months)
3.                                             (9.1%)

Work
1.
0.10
i         0.025           A  P(1  i ) n
4
n  3  4  12              A  5000(1  0.025)12
A  \$6724.44

2.
0.093
i          0.00775        A  P(1  i ) n
12
9000  3000(1  0.00775) n
3  1.00775n
log1.00775 3  n
n  142.3 months  11.9 years
3.
n  18  2  36             A  P(1  i ) n
5 x  x(1  i)36
5  (1  i)36                     4.57% semi-annual = 9.1% annually
1
5  1 i
36

1.0457  1  i  0.0457  4.57%

Page 36                       M30P LP CH 2 & 6 Logs Exp
Appreciation & Depreciation

Examples:

1. A building increases in value by 7% per year. Find its value in 15 years if current value is
\$50 000.                                                                    ANS: 137 951.58

t16 = 50 000(1 + 0.07)16  1             or use      A = 50 000(1 + 0.07)15

2. A building decreases in value by 9% per year. Find its value in 12 years if current value is
\$38 000.                                                                    ANS: \$12 254.07

t13 = 38 000(1 – 0.09)13  1             or use      A = 38 000(1 – 0.09)12

3. Water purifier removes 5% of impurities every hour.

a. What % is removed after 6 hours?                                 ANS: A = 1(1 0.05)6;
A = 0.735; 26.5% is removed
A  P (1  i ) n
A  100(1  0.05) 6
A  73.51%
100  73.51  26.49%

b. How long to remove 75% of impurities?                                   ANS: 27 hours

A  P (1  i ) n
25  100(1  0.05) n
0.25  0.95n
log 0.95 0.25  n
n  27 hours

Assignment
pg. 304 # 1, 6, 13, 15, 20                           or Logs Worksheet E Applications

plus:    Solve for the unknown variables

A                 P                   r        compounding             time
period
1.            \$20 000         \$8000             ?                 semi–annually           7y
2.               3x              x           8.9%/a                  monthly               ?
ANSWERS        1. r  13.5% 2. time  149 months

3. Purifier removes 3% every hour. How long to remove 75% of impurities (0.1)?           ANS: 45.5 h

4. Land valued at \$43 000 is projected to increase by 9%/a. Find its estimated value in 11 years.
(nearest thousand)                                                                ANS: \$111 000

Page 37                     M30P LP CH 2 & 6 Logs Exp
DAY 13

Geometric Series                      Adding of geometric sequences

a(r n  1)
Sn             ,r 1
r 1
rt  a
Sn  n      ,r 1
r 1
(use when you know last term or are finding last term)

Examples:

1. 3, 6, 12, ..., Find:

a. S15    ANS: 98 301                                 b. n if Sn = 25 165 821     ANS: 23

a (r n  1)                                          a (r n  1)
Sn                                                     Sn 
r 1                                                 r 1
3  215  1                                                   3  2n  1
Sn                    98301                           25165821 
2 1                                                           2 1
25165821  3  2n  1
8388607  2n  1
8388607  2n
log 2 8388607  n
n  23

2. 11  12 + ... is a geometric series. Find:

a. S25 (0.0001)                                         ANS: 51.5812

t2 12
r      
t1   11

 12 25 
11        1
S25     11         51.5812
12
1
11

Page 38                        M30P LP CH 2 & 6 Logs Exp
3. 6, 7, ... is a geometric series. Find the smallest value of n so that the sum is at least 1000.
ANS: n = 22

t2 7                     a (r n  1)
r                    Sn 
t1 6                        r 1
 7n 
6   1
1000           
6
7
1
6
500      7n 
 6   1
3      6      
n
250 7
 1
9     6
n
259 7

9     6
259
log 7        n  21.8  22
6 9

4. 5 + 10 + 20 + ... + 1280. Find the sum.                                       ANS: 2 555

We do not know the number of terms, so we need the second formula.
rt  a
Sn  n
r 1
2 1280  5
Sn               2555
2 1

Page 39                         M30P LP CH 2 & 6 Logs Exp
2
5. Bouncing Ball. A ball bounces up                   of its height from which it falls. It is dropped from a height
3
of 81m. Find:

a. height after it hits the ground for 10th time                                           ANS: 1.40m
(54, 36, 24, ...) 54 is first bounce

tn  ar n 1
10 1
2
tn  54               1.4m
3

b. distance travelled until it hits the ground for the eighth time                         ANS: 386.04m

81             54                54         36         36       24    24         16   16       etc.

Bounce 1                            2                      3              4               5    etc.

Obviously there is a geometric sequence. The key is to notice how the 54, 36, 24 etc. are
all doubled. So, calculate the sum, double it and then add the extra 81, which occurs only
once.

a  r n  1
Sn 
r 1
  2 7 
54     1
 3    
S7              152.52m                      152.52  152.52  81  386.04 m
2
1
3

6. You have 2 400 acres of treed land. 5% of the trees are removed each year. How many acres will
be removed in 11 years?                                                  ANS: 1034.88

(trees removed each year: 120, 114, 108.3, ... Find S11
OR # of trees each year: 2400, 2280 (after 1 year = 2nd term), 2166, ... Find t12 and take 2400  t12)

A  P(1  i ) n
A  2400(1  0.05)11
A  2400(0.95)11  1365.12

2400  1365.12  1034.88 acres

Page 40                              M30P LP CH 2 & 6 Logs Exp
a(r n  1)
Sn            ,r 1
r 1
rt  a
Sn  n      ,r 1
r 1

Assignment for day 13
pg. 309 # 1, 5, 6, 22 a, 24 (if you do ex. 6)
plus:

1. 7 + 14 + 28 + ... + 1 835 008. Find the sum.                                           ANS: 3 670 009

rtn  a 2 1835008  7
Sn                           3670009
r 1        2 1

2. 11 + 13 + ... is a geometric series. Find:

a. S13 (0.1) ANS: 470.3                    b. n (smallest) so that Sn  5 000        ANS: n = 27

a  r n  1
  13 13                      Sn 
11    1                              r 1
a  r n  1      11    
Sn                             470.3                         13 n 
11    1
r 1            13
1                                     11     
11                            5000                
13
1
11
  13 n 
909.09  11    1
  11     
           
n
 13 
82.645     1
 11 
n
 13 
83.645   
 11 
log 13 83.645  n  26.49  27
11

Page 41                              M30P LP CH 2 & 6 Logs Exp
3. 5 + 11  ... is a geometric series. Find (0.1):

a. t23       ANS: 170 713 938.7                        b. S24    ANS: 258 204 830.7

t2 11 11                                                  t2 11 11
r                                                       r        
t1 5   5                                                  t1 5   5
tn  ar n 1                                                      a  r n  1
Sn 
 11 
231
r 1
t23   5       
 5                                                      11  24 
5            1
t23  170713938.7                                                    5         
S 24                     258204830.7
11
1
5

4. A bouncing ball bounces up 0.7 of its height from which it fell. It is dropped from a height of
4 000cm. Find:

a. height after it hits the ground for 7th time (0.1)                                             ANS: 329.4cm

tn  ar n 1
tn  2800  0.7 
7 1

tn  329.4cm

b. distance traveled until it hits ground for 7th time (whole number)                              ANS: 20 471cm

4000            2800                   2800      1960    1960     1372 1372             960   960   etc.

Bounce 1                              2                  3                   4               5      etc.

Obviously there is a geometric sequence. The key is to notice how the 2800, 1960, 1372,
960 etc. are all doubled. So, calculate the sum, double it and then add the extra 4000,
which occurs only once.

a  r n  1
Sn 
r 1
2800  0.76  1
S6                              8235.276cm               8235.276  8235.276  4000  20471cm
0.7  1

Page 42                              M30P LP CH 2 & 6 Logs Exp
DAY 14             (Short Day)

Sigma Notation (                  )

Examples:

1. Expand and evaluate (Find sum).

7
a.     5k  k
k 3
2
= 24 + 36 + 50 + 66 + 84 = 260

This means put 3, 4, 5, 6 and 7 into the formula 5k  k 2 and add your answers.

25
b.     5(2
k 2
k 1
)                                             ANS: 167 772 150 (10 + 20 + 30 + ...)

* a is not necessarily 5, but r is definitely 2 (if coefficient of k is 1),
# of terms = 25  2 + 1 = 24

38
c.     2(1.1)
k 5
k 3
ANS: 594.05    (2.42 + 2.662 + 2.9282 + ...)

* a = 2.42, r = 1.1, n = 38–5+1=34

2. Write using                     notation.

17
a. 4 + 8 + 16 + ... + 262 144                                                            ANS:    4(2)
k 1
k 1

* Find n first (17) then use k = 1 to k = 17, a = 4, r = 2.

12
b. 5  15  45  ... 885 735                                                           ANS:    5(3)
k 1
k 1

k 1
1                                                                 12
1              12
c. 27 + 9 + 3 + ... +                                                              ANS:  27   or          3      k4

6561                                                              k 1 3             k 1

21
3. How many terms in the series:                 
k  p 3
4(5) k 1                        ANS: 19  p

Page 43                           M30P LP CH 2 & 6 Logs Exp
Assignment
pg. 309 # 8, 16
plus:

1. Find sum:

16                                                   19
a.    3(2)k 1
k 2
ANS: 196 602                   b.   100(0.6)
k 5
k 3
ANS: 89.96

a  3 2
2 1
6                              a  100(0.6)53  36
t2  100(0.6) 63  21.6
t2  3  2 
31
 12
21.6
r2                                                  r         0.6
36
n  16  2  1  15
n  19  5  1  15
6  215  1
S15                       196602                           36  0.615  1
2 1                                      S15                       89.96
0.6  1

2. Write using               notation.

12
a. 2 + 10 + 50 + ... + 97 656 250                                                  ANS:       2(5)
k 1
k 1

n 1
tn  ar
97656250  2(5) n 1
48828125  5n 1
log 5 48828125  n  1
n  12

k 1
9
3
b. 64 + 96 + 144 + ... + 1 640.25                                                  ANS:       64  2 
k 1  
tn  ar n 1
1640.25  64(1.5) n 1
25.629  1.5n 1
log1.5 25.629  n  1
n9

Page 44                                M30P LP CH 2 & 6 Logs Exp
Decibel records
1976
The Who were listed as the "record holder", at 126 dB, measured at a distance of 32 meters from the speakers at a
concert at The Valley (home ground of Charlton Athletic F.C.) on 31 May 1976.[2]
1984 and 1994
The heavy metal band Manowar is one claimant of the title of "loudest band in the world",[5] citing a measurement
of 129.5 dB in 1994 in Hanover.[6] However, Guinness Book of World Records listed Manowar as the record holder
for the loudest musical performance for an earlier performance in 1984. Guinness does not recognize Manowar's
later claim, because it no longer includes a category of loudest band, reportedly because it does not want to
encourage hearing damage. Manowar broke their own record again in 2008 at the Magic Circle Fest with 139dB.

1997

Metallica has styled itself the "loudest band in the world". However, after one concert on 11 November 1997,
which the band dubbed the "Million Decibel March", the Philadelphia Inquirer reported that "neighbors who [had]
feared the worst from the self-styled Loudest Band in the World complained more about the sound from the news
2007
British punk band Gallows allegedly broke Manowar's previous record for loudest band in the world, claiming to
have achieved 132.5 dB; however, this record was claimed in an isolated studio environment as opposed to live.
Other dates
Other previous record holders include Deep Purple (117 decibels), Iron Maiden (1985 Guinness World Book of
Records), The Rolling Stones, Motörhead, AC/DC, My Bloody Valentine (132 decibels), and KISS (120 decibels).
Although it has never been made official, some reports suggest that New York noise rock band, Swans reached 140
decibels at some gigs.

Environmental Noise

Whisper Quiet Library                                           30dB
Normal conversation (3-5')                                       60-70dB
Telephone dial tone                                            80dB
City Traffic (inside car)                                        85dB
Train whistle at 500', Truck Traffic                                  90dB
Subway train at 200'                                           95dB
Level at which sustained exposure may result in hearing loss                     90 - 95dB
Power mower at 3'                                            107dB
Snowmobile, Motorcycle                                          100dB
Power saw at 3'                                             110dB
Sandblasting, Loud Rock Concert                                      115dB
Pain begins                                               125dB
Pneumatic riveter at 4'                                        125dB
Even short term exposure can cause permanent damage -
140dB
Loudest recommended exposure WITH hearing protection
Jet engine at 100', Gun Blast                                      140dB
Death of hearing tissue                                         180dB
Loudest sound possible                                          194dB

Page 45                    M30P LP CH 2 & 6 Logs Exp

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