# Exponential Growth

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```					Exponential Growth
Laws of Exponents and Geometric Patterns
Laws of Exponents
Consider the expression       The ‘2’ is called the coefficient
2x4 = 2 (x)(x)(x)(x)          and the x4 is called the power. ‘x’
is the base and 4 is the exponent
With this we can get the Laws of exponents
From this we see that
Multiplying Powers with the same base:     when multiplying powers
with the same base, we
keep the base and add the
exponents.
(a5)(a2) = (aaaaa)(aa)

(a5)(a2) =aaaaaaa
(a5)(a2) =a7
Laws of Exponents
Recall that 2x4 = 2(x)(x)(x)(x)

With this we can get the Laws of exponents
From this we see that
Multiplying Powers with the same base:    when multiplying powers
with the same base, we
keep the base and add the
exponents.

From this we see that when
Dividing Powers with the same base:
dividing powers with the
b7                                  same base, we keep the
b4                                  base and subtract the
bbbbbbb                           exponents.

bbb
 bbbb
 b4
Laws of Exponents
Recall that 2x4 = 2(x)(x)(x)(x)

With this we can get the Laws of exponents
From this we see that
Multiplying Powers with the same base:    when multiplying powers
with the same base, we
keep the base and add the
exponents.

From this we see that when
Dividing Powers with the same base:
dividing powers with the
same base, we keep the
base and subtract the
exponents.

Power of a Power       When we have a power of a
power, we keep the base
( w3 ) 2  ( www) 2  ( www)( www)  w6      and multiply the exponents.
Using the Laws of Exponents
Simplify the following expressions

a) (w3)(w10)         b) (4d2)(10d6)       c) 15 s 8
3s 6
  2 
4
e)  s t 
3     9              5 3                     2
d) ( 2 )( 2 )            3              f) 23        5

     
2                 st                2 2 
4       2 2

a) w13                b) 40d8

c) 5s2                d) 211

e) s8t8               f) 23
Zero and Negative Exponents
3
What does 2 equal?
23
We also know that
23 = 2x2x2 = 8        When dividing powers with
the same base, we keep the
23 8                 base and subtract the
3
 1              exponents. So…
2 8
23             So 20 must equal 1.
 20        In fact, ANYTHING
23
to the zero equals 1.

b0 = 1
Zero and Negative Exponents
2
What does 3 equal?
33
We also know that
32 = 3x3 = 9                       When dividing powers with
33 = 3x3x3=27                      the same base, we keep the
base and subtract the
32 9             1                    exponents. So…
    0.3333 
33 27            3                  32           So 3-1 must equal 1/3.
 31
33
A negative exponent means
you flip the base and keep the
positive exponent
Zero and Negative Exponents
Ex1. Simplify the following leaving no negative exponents.

b) q 
3 0
a) 53                                                2 a b
c) x
 
53
4
q 
3 2
x 3 a b

Answers: a) 5 9  1                 b) q 6                  c) x  a  2b
9
5

Ex2. Evaluate (which means “get the value of the expression”)
2
a) 24      b)  3 
           c) 2 2       d)  30
2
Answers: a) 1            b) 4         c) 1        d) -1
16           9             4
Zero and Negative Exponents
Ex 3. Evaluate each expression

2                                                3
2                            2            9          1
3
2
a.                          3                       b.    53  125
3                     2
2  2                4          5
2
 1     
2

c.    7 
 2 

0




d. 2  61

1 2

     1 2
2 1                                                            
2             1
1                                   2 2

2

2                                             2    6
  
 6  6 
3
5                    1 2
 4  1
2                                  3

   
3 2

 6 
1
                     4 2              2
 52                           25
    9
4
Changing the base
The laws of exponents only work when there’s
a common base. Sometimes we can change
the base of a power to make it common to
another.

Ex. Simplify         4 8 
3x       x
All of these bases can be
16  2           written as a base of 2.

2  (2 ) 
2 3x

4 2
3 x
This doesn’t look like
(2 )                          simplifying….

2 2 
6x        3x            Ok maybe it does…

2 8
2  9x

29 x 8
Yep, that’s simpler. It is
now expressed as a single
8
2                                 base.
Changing the base
Ex 2. Evaluate without a calculator

27 81 
2         5
This expression has some
27  33
7             nasty values! You are
9  1 
4
                           NOT required to know                81  34
3                        81-5 or (1/3)7. BUT…
Power of a                                                                           9  32
Power =
keep the        3  3 
3 2      4 5                    All of these bases can be
written as a base of 3.
1 1
3
(3 ) 3 
base and          2 4      1 7                                                     3
multiply the
exponents
3 3 
6       20
Multiplying
powers with the
3 
14        Dividing powers
with the same
(315 )
(3 )3 
8       7         same base = keep                   base = keep the
the base and add                   base and subtract
the exponents                      the exponents
314( 15)
31 
Solving Exponential Equations
When solving exponential equations (right now) we will
look to get a common base.

If the bases are equal, then
the exponents must be equal

Ex 1. Solve for ‘x’      or Find the Root of the equation

2 x 4
We must
9             27   4
recognize that
these bases can
be written using
base 3.

3 
2 2 x 4
 
 3   3 4
Power of a
power = keep
4x  8  12

34 x8  312
the base and
multiply the
4x  12  8
exponents
x 5
Solving Exponential Equations
Ex 2. Solve for x.

64 8 
3x      4x

1
128
Recognize the base of 2

Power of a
2  2 
6 3x     3 4x
 2 7
power = keep
the base and
multiply the

2 2   2
18 x    12 x       7
exponents

230 x  27            If the bases are equal, then
the exponents must be equal
30 x  7
x   30
7
Solving Exponential Equations
Ex 3. Find the root(s) of the equation below.

3(8 x )  48  0           ISOLATE THE POWER!!

3(8 x )  48

(8 x )  16           Base of 2

23 x  24

3x  4
x       4
3
Solving Exponential Equations
Find the root(s) of the equation below.

2 x2
a. 2             7  71                 b. 4(2 x 3 )  256

d . x  3 x  10  0
2
x 2
c. 4          8   4
Solving Exponential Equations
Find the root(s) of the equation below.

2 x2
a. 2            7  71                  b. 4(2 x 3 )  256
2 2 x  2  64                          2 2 ( 2 x  3 )  28
22 x 2  26                               2 x5  28
2x  2  6                                  x5 8
2x  4                                      x3

x2
Solving Exponential Equations
Find the root(s) of the equation below.

d . x  3 x  10  0
2

c. 4 x 2  84
2 
2 x 2
 
 2  3 4
( x  5)(x  2)  0

22 x 4  212                    ( x  5)  0 ( x  2)  0
2 x  4  12                            x 5       x  2
2 x  16
x 8
Patterns (again)
We have seen that a pattern can be represented as an equation.

Linear Pattern:    t n  t1  (n  1)d      y  mx  b
CD is on D1

Quadratic Pattern:    t n  an 2  bn  c        y  ax 2  bx  c
CD is on D2

But what about a pattern like this?

3,6,12,24,48
Geometric Patterns
We can quickly
3, 6, 12, 24, 48                                We can divide the
see that there is
terms to find a
no common
COMMON RATIO
difference for this
3 6 12 24            pattern.                Patterns with a COMMON
RATIO are called Geometric
Patterns
So we can express this
pattern as:                                      3, 6, 12, 24, 48

3
3x2                                              2    2   2 2
3x2x2                   Or…       3 x 20
3x2x2x2                           3 x 21
3x2x2x2x2                         3 x 22              Remember: 20 = 1
3 x 23
3 x 24
Geometric Patterns
3, 6, 12, 24, 48

3 x 20
3 x 21               We see from this that this
3 x 22               geometric pattern can be
3 x 23               represented by the equation:
3 x 24
t n  3(2) n 1     t 4  3(2) 41
Let’s check: If n = 4
t 4  3(2)3
The 2 is the common ratio of                             t 4  3(8)
the pattern and is the base in the
equation.                                                t 4  24

The 3 is the first term of the pattern and is the
coefficient in the equation.
Geometric Patterns
In general, a geometric pattern can be written
using the equation
n 1      Where t1 is the first
t n  t1 (r )             term of the pattern
(when n=1) and where
‘r’ is the common ratio.

The pattern 5, 10, 20, 40, 80,… can be
represented by the equation
We can check by
tn  t1 (r ) n1               plugging in n =5

tn  5(2)   n 1
t5  5(2) 51
t5  5(2) 4 t5  5(16 )
t5  80
Geometric Patterns
Find the 10th term in each pattern:

a) 100, 50, 25, 12.5,…
a) CR = ½               t1 = 100
b) 0.25, 1,4, 16,…
n 1
c) 5, 8, 11, 14, …                              1
tn  100 
d) 1, -2, 4, -8, 16, …                          2
10 1
1
t10  100 
2
9
1
t10  100 
2
t10  100(0.001953125)
t10  0.1953125
Geometric Patterns
Find the 10th term in each pattern:

a) 100, 50, 25, 12.5,…
b) CR = 4             t1 = 0.25
b) 0.25, 1,4, 16,…

c) 5, 8, 11, 14, …                      tn  0.254
n1

d) 1, -2, 4, -8, 16, …
t10  0.254 
10 1

t10  0.254 
9

t10  0.25(262144)
t10  65536
Geometric Patterns
Find the 10th term in each pattern:

a) 100, 50, 25, 12.5,…
c) CD = 3          t1 = 5
b) 0.25, 1,4, 16,…
tn  t1  (n  1)d
c) 5, 8, 11, 14, …
tn  5  (n  1)3
d) 1, -2, 4, -8, 16, …

t10  5  (10  1)3
t10  5  (9)(3)
t10  32
Geometric Patterns
Find the 10th term in each pattern:

a) 100, 50, 25, 12.5,…
d) CR = -2            t1 = 1
b) 0.25, 1,4, 16,…
tn  1 2
n 1
c) 5, 8, 11, 14, …

d) 1, -2, 4, -8, 16, …
t10   2
101

t10   2
9

t10  512
Geometric Patterns
For the pattern below, which term has a value of 768?

6,12,24,48, …

tn  62
n 1

768  62 
n 1

128  2 
n 1
7  n 1

2 7  2 
n 1                          n 8
The 8th term has a
value of 768.
May I Have Another Word (Problem)
A great many things in nature grow exponentially. Each of
these situations can be modeled with a geometric pattern and
thus an exponential equation.
Ex. A certain type of bacteria doubles every
six hours. The experiment begins with 1
bacteria.                         Yet another 6
Another 6                      hours later…
hours later…
6 hours later…
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every
six hours. The experiment begins with 1
bacteria.
After a short
while…
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every       We can set up a pattern
six hours. The experiment begins with 1            to show the number of
bacteria.                                          bacteria.
Number of hours since the              0       6      12     18
experiment began
Number of bacteria present             1       2      4      8

We see there’s
a CR of 2 in the
pattern.
However, this pattern
involves ‘n’ values that do
not increase by 1. They
increase by 6.
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every       We can set up a pattern
six hours. The experiment begins with 1            to show the number of
bacteria.                                          bacteria.
Number of hours since the             0       6      12     18
experiment began
Number of bacteria present            1       2      4      8

When this happens, the                         We see there’s
equation must change.                          a CR of 2 in the
pattern.

Notice as well that we do                     However, this pattern
not know the t1 value. We                     involves ‘n’ values that do
know t0 and t6 instead.                       not increase by 1. They
increase by 6.
We cannot use
t n  t1 (r ) n 1
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every       We can set up a pattern
six hours. The experiment begins with 1            to show the number of
bacteria.                                          bacteria.
Number of hours since the             0       6      12    18
experiment began
Number of bacteria present            1       2      4     8

We’ll need to use the equation:
x
y  A0 (r )     period           Where y is the amount at
time x and A0 is the
So we get the                                         original amount and the
function                                              period is the amount by
x                       which the x values
y  1(2)     6                       increase.
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every
six hours. The experiment begins with 1
bacteria. How many bacteria are present
after 50 hours?

Just from the question we can see
that r = 2 (doubles) and t0 = 1so we
can get the equation
x
y  1(2)     6
Now let x = 50 and
solve for y.

50
y  1(2)     6

There would be 322
y  322.5398              bacteria at t = 50
hours.
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every
six hours. The experiment begins with 1
bacteria. When will there be 128 bacteria
present?
x
y  1(2)     6       Now let y = 128 and
solve for x.
x

128  1(2)          6

x
2 2
7       6

x
7
6                       At t=42 hours there
x  42                      will be 128 bacteria.

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