# Lesson 5 Exponents and Logarithms

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```					                                  Lesson 5
Exponents and Logarithms
PURPOSE
This lesson concentrates on simplifying expressions containing exponents and working with
exponential functions. This also leads to the inverse of the exponential function, the logarithmic
function.

OBJECTIVES
After completing this lesson, you should be able to:

simplify expressions containing negative exponents;

simplify expressions containing fractional exponents;

define and use exponential functions;

define and apply the natural exponential function;

define and apply logarithms;

apply laws of logarithms; and

solve exponential equations and change logarithms from one base to another.

Textbook, Chapter 5, Sections 5-1 through 5-7 (pages 169–207)

COMMENTARY
Section 5-1: Growth and Decay: Integral Exponents pp. 169–175

Let's review work with exponents.

1. b x + b x = 2 b x

2. b x•b y = b x + y

3. 3 2×3 5 = 3 7
Lesson 5

4. (b x) y = b x y
¡ ¢5
5. 3 2 = 3 10

bx
6.   by
= bx− y

35
7.        = 3 5 − 2 = 3 3 = 27
32

32
8.        = 3 2 − 5 = 3 −3 =     1
35                          27

32
9.        = 32−2 = 30 = 1
32

This last problem showed that 3 0 = 1. It is a fact that b 0 = 1. In problem 8, the simplification has a
negative exponent. That is because there were more 3s in the denominator than in the numerator. For
that reason, the end result has the 27 in the denominator. If you compare the results of problems 7
and 8, you see they are the inverse of each other. So b −3 is the inverse of b 3. The inverse of b 3is 13 .
b
Thus b −3 =     1
and     1
= b 3. When a problem is completely simplified, there are no negative
b3         b −3
exponents or powers of powers.
¡           ¢ −1
Example 1: Simplify 3 −2 + 3 2      .

You cannot add unless exponents and bases are identical. So simplify 3 −2 and 3 2 first:
µ       ¶ −1       µ        ¶ −1
1                  82              9
+9       =                   =      .
9                  9               82
³           ´ −2 ³ 4 ´ −1
a −3         a
Example 2: Simplify                                   .
b −2         b 3

When raising a power by a power, multiply the exponents:

a 6 a −4
×     .
b 4 b −3

a2
.
b
¡                ¢
Example 3: Simplify a b −3 a b 2 − a −3 b 2 .

Distribute the monomial:

Ia b −3M Ia b 2M − Ia b −3M Ia −3 b 2M.

a 2 b −1 − a −2 b −1.
Lesson 5

Invert terms with negative exponents:

a2   1
− 2 .
b   a b

a4   1  a4 − 1
− 2 = 2 .
a2 b a b  a b

J−4 a −4N
2
Example 4: Simplify               −1
.
−4 Áa 8˜
È ˘

Simplify the numerator by squaring −4 and multiplying the exponents:

16 a −8
¡ ¢ −1 .
−4 a 8

Simplify the denominator:

16 a −8
.
−4 a −8

Divide and subtract the exponents:

−4 a −8 − (−8) ⇒ − 4 a 0 ⇒ − 4.

Exponents are used to find values that increase or decrease exponentially. This occurs when an item
increases or decreases by a set percentage every year. If the cost of a car increases yearly by 8%,
next year it would cost 100% of this year's cost plus an additional 8%. Then the year after that, it
would increase by another 8%. So every year the new price could be found by multiplying by 1.08,
the decimal for 108%. A function can be used to find the new price in t years:

A (t) = A 0 (1 + r) t.

A 0 is the beginning amount, r is the rate of increase as a decimal, and t is the number of years the
price has been increasing. This function can also be used to find the price years in the past, if a
negative t is used.

Example 5: A car now costs \$4,000. Car prices are increasing at 10% a year. (a) What will the price
be in 6 years? (b) What was the price 5 years ago?

(a)   A (6) = 4, 000 (1 + .10) 6
A (6) = 4, 000 (1.10) 6
A (6) = 4, 000 (1.771561)
A (6) = \$ 7, 086.24
Lesson 5

(b) A (−5) = 4, 000 (1 + .10) −5

A (−5) = 4, 000 (1.10) −5
A (−5) = 4, 000 (.621)
A (−5) = \$ 2, 484

Example 6: Let's try problem 38 on page 174 of your text.

(a)
4 9×8 −4
16 3

Change all bases to powers of 2:
¡ 2¢ 9 ¡ 3¢ −4
2 × 2
¡ 4¢ 3
2

2 18×2 −12
2 12
26           1
12
⇒ 2 −6 ⇒ .
2            64

(b)
37 5
p ×9
27 12

Change all bases to powers of 3:
¡ ¢5
3 7× 3 2   3 7×3 10
q¡ ¢ ⇒ √            .
12
33        3 36

The square root of a number is also a power of 1 :
2

√      1
b = b 2.

3 7×3 10  3 17         1
¡ 36¢ 1  ⇒ 18 ⇒ 3 −1 ⇒
3            3
3 2

(c) s
Í Á n 4 n˜ ¸
Î
3 Î È125 •5 ˘
Î
Î
Î 25 −n
Î
Î
Ï

3" ¡¡ ¢
Change all bases to powers of 5:
¢#  sÍ
n
5 3 •5 4 n       Î 7n ¸
3 Î 5
3
¡ 2¢ −n       ⇒ Î −2 n .
Î
Î
Î
Î
5              Ï5
Î
Lesson 5

The cube root of a number is also a power of 1 :
3
q
3          1
b = b 3.
Í 7 n ¸1/3
Î5
Î
Î            57 n/3
Î
Î −2 n     ⇒ −2 n / 3 ⇒ 5 7 n / 3 − (−2 n / 3) ⇒ 5 9 n / 3 ⇒ 5 3 n
Î5
Î           5
Ï
Î

Example 7: Let's try problem 48 on page 175 of your text:

2 2 x − 3•2 x + 1 + 8 = 0.

If 2 x + 1was 2 x, the problem could be factored. If I divide 2 x + 1by 2 , it will be 2 x:

2 2 x − 3 (2)•2 x + 8 = 0
(2 x) 2 − 3 (2)•2 x + 8 = 0.

Replace 2 x with y and factor:

y2 − 6 y + 8 = 0
(y − 4) (y − 2) = 0
y = 4;       y = 2.

Replace y with 2 x:

2 x = 4;       2 x = 2;
x = 2;       x = 1.

Study Exercises
Complete odd-numbered problems 1–49 in the Written Exercises on pages 173–175 of your text.

Section 5-2: Growth and Decay: Rational Exponents pp. 175–180

As briefly mentioned in the previous section, roots are fractional exponents. Other than that, all
previous rules hold.
q¡      ¢
Example 1: Write      4 x 3 in exponential form:

4 x 3 ⇒ I4 x 3M ⇒ 2 x 3 / 2.
q¡     ¢         1/2

¡           ¢4
Example 2: Simplify 64 − 1 / 16 :

I64 − 1 / 16M ⇒ 64 − 1 / 4 ⇒ I64 1 / 4M
−1       √ −1               1
= Á 64˜ ⇒ 2.828 −1 ⇒
4                                       4
.
È   ˘              2.828
Lesson 5

Example 3: Let's try problem 26 on page 178 of your text:

4 a b−1/2 − 2 a b1/2
¡    ¢−1/2
a2 b

4 a b−1/2       2 a b1/2
− −1 −1 / 2
a −1 b − 1 / 2 a b
4 a 2 − 2 a 2 b.

Example 4: Solve 4 x = 2 20:

I2 2M = 2 20.
x

They have the same bases so the exponents must be the same:

2 2 x = 2 20
2 x = 20
x = 10.

Example 5: Solve 3 (x − 2) −2 = 27:

3 (x − 2) −2 = 27
(x − 2) −2 = 9.

Raise both sides by a power of − 1 :
2

x − 2 = 9−1/2
1
x−2=
3
7
x= .
3

Example 6: Let's try problem 52 on page 179 of your text.

There is a greatest common factor of (2 x + 1) − 1 / 3:

(2 x + 1) − 1 / 3 (2 x + 1 − 4)
(2 x + 1) − 1 / 3 (2 x − 3).

Example 7: Let's try problem 38 on page 179 of your text:

A (t) = A 0 (1 + r) t.

Divide by 140:

182 = 140 (1 + r) 4
1.3 = (1 + r) 4.
Lesson 5

Raise both sides to the power of 1 :
4

(1.3) 1 / 4 = 1 + r
(1.3) 1 / 4 − 1 = r
1.067 − 1 = r
.067 = r.

The annual percent of increase has been 6.7 %.

Study Exercises
Complete odd-numbered problems 1–57 in the Written Exercises sections on pages 178–180 of your

Section 5-3: Exponential Functions pp. 180–186

The general form for an exponential function is f (x) = a b x. With this form, function values can be
given and the function can be found. Given f (0) = 4 and f (3) = 32, the function can be found using
substitution:

f (0) = a b 0 = 4;

a = 4 since b 0 = 1;

f (3) = 4 b 3 = 32
b3 = 8
b=2
f (x) = 4 (2 x).

We have used A (t) = A 0 (1 + r) t for exponential growth and decay when the rate of change is known.
If the rate is not known, A (t) = A 0 b t / k can be used. k is the time needed to multiply A 0by b.

Example 1: Let's try problem 8a on page 183 of your text:
µ ¶ 3, 200 / 1, 600
1
A (t) = 1                     .
2

1
The value of b is   2
because the half-life is the time at which half of the substance will remain. The
value of k is 1,600 because it takes 1,600 years for 1 of it to remain. The value of t is 3,200 because
2
we want to see how much remains after 3,200 years:
µ ¶2
1
A (t) =
2
1
A (t) = .
4
1
After 3,200 years   4
of a kilogram will remain.
Lesson 5

Example 2: Let's try problem 16 on page 184 of your text:
µ ¶ 30 / 3
1
A (t) = A 0
2
µ ¶ 10
1
A (t) = A 0
2
A (t) = A 0 (.000977).

There will be .000977 times the original amount of medicine in the blood stream after 30 days.

Example 3: Let's try problem 12 on page 184:

(a) A (10) = 30, 000, 000 (1 + .03) 10

A (10) = 30, 000, 000 (1.03) 10
A (10) = 40, 317, 491.

The population in 2,000 will be 40 million.

(b) The rule of 72 states an estimate for doubling time can be found by dividing 72 by the rate.
So 72 = 24. It would take approximately 24 years to double.
3

Example 4: One last problem: problem 28 on page 186 of your text.

(a) The population doubled from 100 million to 200 million in 60 years. So the doubling time is
£          ¤
60 years. Using A (t) = A 0 b t / k , A (t) = 200 (2) t / 60 .

(b) After trying 30 years, 36 years, 33 years, and 35 years as t values in the equation above, I
have concluded that t = 35 years gives a population of 299.66 million. This is very close to
300 million.

Study Exercises
Complete odd-numbered problems 1–29 in the Written Exercises on pages 183–186 of your text.

Section 5-4: The Number e and the Function e x pp. 186–190

The letter e has a special meaning in math, statistics, and physics. It is the value of the natural
exponential function at 1. The letter e was chosen for this value in honor of its discoverer, Leonhard
¡       ¢x
Euler. Its value can be approximated using y = 1 + 1 . As x gets increasingly larger, the value of y
x
approaches 2.7182818. This process of placing larger and larger values in for x can be shown using
¡      ¢n
calculus terminology. The limit lim 1 + 1 terminology means you can keep placing values in for
n
n→∞
n until you reach infinity.

The value of e is used in this section to find annual yield on a savings account that has interest
compounded daily. The function P (t) = P 0 e r t can be used for any item that grows continually. Your
calculator should have an e x key to help you work with this function.
Lesson 5

Example 1: Suppose you invest \$1.00 at 8% interest. How much would you have at the end of the
year if interest compounds (a) quarterly, (b) monthly, (c) continuously?

(a)
Quarterly means four times a year. So the 8% interest will be paid in 2% increments. You
would have 102% of your money every 3 months:

(1.00) (1.02) 4
(1.00) (1.0824) = \$ 1.0824.

(b)
Monthly means 12 times a year. So the 8% interest will be paid in .667% increments. You
would have 100.667% of your money every 12 months:

(1.00) (1.00667) 12
(1.00) (1.0830) = \$ 1.083.

(c)
For continuous compounding, use P (t) = P 0 e r t. The value of r is the interest rate as a
decimal, and t is the time in years:

P (1) = (1.00) e (.08) (1)
P (1) = \$ 1.0833.

Example 2: Let's try problem 12 on page 189 of your text.

(a)
The problem states t represents number of days from now. So to find the number of ladybugs
present now, t = 0.

A (0) = 3000 e (.01) (0)     e0 = 1
A (0) = 3000

(b)
7 days in a week:

A (7) = 3, 000 e (.01) (7)
A (7) = 3, 000 e (.07)
A (7) = (3, 000) (1.0725)
A (7) = 3, 217.5.

In a week, there will be 3,217 ladybugs.
Lesson 5

Example 3: Let's try parts (a) and (c) in problem 20 on page 190 of your text.

(a)
When h = 1, point A is (0, f (0))and point B is (1, f (1)).

f (0) = e 0 = 1
f (1) = e 1 = 2.7182818
A (0,   1)       B (1,   2.7182818)
y 2 − y 1 1 − 2.7182818
Slope =            =              = 1.7182818
x2 − x1       0−1

(c)
When h = .01, point A is (0, f (0))and point B is (.01, f (.01)).

f (0) = e 0 = 1
f (.01) = e .01 = 1.01005
A (0,   1)       B (.01,   1.01005)
y 2 − y 1 1 − 1.01005
Slope =             =            = 1.005
x2 − x1     0 − .01

Study Exercises
Complete odd-numbered problems 1–23 in the Written Exercises section on pages 189–190 of your

Section 5-5: Logarithmic Functions pp. 191–197

If I needed to solve the problem 18 = 2 x, I would need some way to solve for x. Normally I would
use the inverse function. For example, let's examine how we solve 2 x + 3 = 13.

The inverse of addition is subtraction:

2 x + 3 = 13
.
− 3   − 3
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
The inverse of multiplication is division:

2 x + 3 = 13
− 3   − 3
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 x 10
=
2    2
x = 5.

For the problem 18 = 2 x, however, the four basic operations will not get the x alone. I need an
inverse function for raising to a power. That inverse function is the logarithmic function. A
logarithmic function is the inverse of an exponential function. By definition, x = b a is log b x = a.
Lesson 5

Notice the exponent, a, is alone in the log function. In log b x = a, the b is called the base. Even
x = e a can be written as log e x = a. Not surprisingly, the log of base e has been given a special name,
the natural logarithm function. And it has a special look, log e x = a ⇒ ln x = a. The log base e is
replaced with natural log, ln. Let's do a little work with converting from log form to exponential
form.

Example 1: Write log 3 81 = 4 in exponential form:

3 4 = 81.
1
Example 2: Write log 5 125 in exponential form.

1
We are missing the value of the log. I will set log 5 125 equal to x:

1
log 5       =x
125
1
5x =        .
125

Now I will take it a step further and determine the value of x:

5 x = (125) −1

5 x = I5 3M
−1

5 x = 5 −3
x = − 3.
1
So the value of the log, −3 , is the exponent of 5 (the base). That gives you    125
.
√
Example 3: Write log 6 3 (1 / 6) in exponential form and find its value.

Write in exponential form:
sµ ¶
3 1
6x =      .
6

The cube root is the power of 1 :
3

µ ¶1/3
1
6 =
x
6

6 x = I6 −1M
1/3

6x = 6−1/3
1
x= − .
3
q¡ ¢
So the value of the log, − 1 , is the exponent of 6 (the base). That gives you
3
3 1 .
6
Lesson 5

Example 4: Let's try problem 18 on page 195 of your text.

(a)
When a base is not given, the is base 10. log 10 10 8 = x.

log 10 8
10 x = 10 8
x = 8

(b)
log 2 2 8 ⇒ log 2 2 8 = x
2x = 28
x=8

It looks like when the base is raised to a power, the power is the value of the log: log b b x = x.

(c)
log 5 5 8

So this would have a value of 8.

(d)
ln e 8

Remember ln is log e.

log e e 8 = 8

Example 5: Find 3 log 4 16.

Start with log 4 16. That has a value of 2. So now find 3 2 = 9.

Example 6: Find 4 log 4 64.

Again, log 4 64 = 3, so 4 3 = 64. Notice the 64 in the problem. So b log b x = x.

Example 7: If f (x) = 5 x, what is f −1 (x)?

y = 5x

Swap the x and y:

x = 5 y.

Solve for y using log function; write in log form:

log 5 x = y
f −1 (x) = log 5 x.
Lesson 5

Example 8: Let's try problem 34 on page 195 of your text.

(a)
y = log 2 (x − 2)

The domain involves values placed in for x. You can never take the log of a negative number
or zero, because raising a number to a power will never make the number zero or negative. So
x − 2 > 0. The domain is {x| x > 2}. The range involves the values of y. Keeping in mind the
value of a log is the power the base is raised to, y can be any number. You can raise numbers
to negative powers, fractional powers, positive powers, and zero. The range is all real
numbers. The zeros would be found like this:

0 = log 2 (x − 2).

Rewrite in exponential form:

20 = x − 2
1 = x−2
3 = x.

The zero is 3.

(b)
y = log 2 x − 2

Here you are taking the log of x only. The domain is {x| x > 0}. The range is all real numbers.
Zeros are found in this way:

0 = log 2 x − 2.

Rewrite in exponential form:

2 = log 2 x
22 = x
4=x

The zero is 4.

¡       ¢
Example 9: Solve for x: log x 2 − 1 = 2.

Write in exponential form; this log has a base of 10:

10 2 = x 2 − 1
101 = x 2
√
± 101 = x.

Example 10: Solve for x:

log 6 (log 2 x) = 1.
Lesson 5

Let's replace log 2 xwith y:

log 6 y = 1.

Write in exponential form:

6 1 = y.

So now we know log 2 x = 6.

Rewrite log 2 x = 6 in exponential form:

26 = x
64 = x.

Example 11: Solve ln (x − 2) = 1.

log e (x − 2) = 1.

Write in exponential form:

e1 = x − 2
e + 2 = x.

Example 12: Solve ln x = − 0.52 using your calculator.

Written in exponential form:

e −0.52 = x
.5945 = x.

Study Exercises
Complete odd-numbered problems 1–49 in the Written Exercises section on pages 194–197 of your

Section 5-6: Laws of Logarithms pp. 197–203

Let's experiment with log 2 4 + log 2 8 = log 2 32. Is it a true statement?

log 2 4 = 2

log 2 8 = 3

log 2 32 = 5

2+3 = 5

So it would appear that to add logs of like bases, you multiply numbers you are taking logs of. This
¡ ¢¡ ¢
is kind of the inverse of x 3 x 4 = x 7. It is a multiplication problem, and you add exponents. So in
x4
log 2 4 − log 2 8, would you divide because in        you subtract exponents?
x8
Lesson 5

1
log 2 4 − log 2 8 = log 2
2
2−3 = −1
1
2 −1 =
2

Yes, you divide the 4 and 8 to get an equivalent log for the difference.

What about log 2 4 3? That would be log 2 64, which is 6. That is three times the value of log 2 4, 2. So
log 2 4 3 could be written as 3 log 2 4. These observations are written as laws on page 197 of your text.
These laws can be used to simplify log problems and make work easier.

Example 1: Let's try problem 4 on page 200 of your text:
√
4
log M N
log M (N) 1 / 4.

Using law number 1:

log M + log N 1 / 4.

Using law number 4:

1
log M +      log N
4

or

log N
log M +          .
4
√
Example 2: Rewrite log     (1 / x) in terms of log x:
sµ ¶       µ ¶1/2
1         1
log      = log        .
x         x

Using law number 4:

= log x − 1 / 2
−1
=       log x
2

or

−log x
=          .
2
Lesson 5

Example 3: Let's try problem 10 on page 200 of your text:

log 2 48 − (1 / 3) log 2 27.

Using law number 4:

log 2 48 − log 2 27 1 / 3.
√
log 2 48 − log 2 3         27 1 / 3 =
3
27 = 3

Using law number 2:
µ        ¶
48
log 2
3
log 2 16.

Example 4: Let's try problem 16 on page 200 of your text:

1
(log b M + log b N − log b P).
2

Using law number 1:

1
(log b M N − log b P).
2

Using law number 2:

1
(log b M N / P).
2

Using law number 4:
p
log b    (M N / P).
√
Example 5: Simplify e ln x.
√                             √
Let's work with ln x. That is, e to some power is x. Now in our original problem, e is taken to
√
that power. So it now has a value of x.

Example 6: Simplify 10 1 + 2 log x.

Let's simplify 1 + 2 log x as one log. The base is 10 so log 10 = 1.

1 + 2 log x = log 10 + 2 log x
= log 10 + log x 2

This means 10 to some power is 10 x 2:

= log 10 x 2.

Now back to the original problem:

10 1 + 2 log x.
Lesson 5

Now 10 is taken to the power mentioned above. So it has a value of 10 x 2.

Example 7: Express in terms of x, ln y = 2 ln x − ln 6:

ln y = 2 ln x − ln 6
x2
ln y = ln
6
x2
y =          .
6

Example 8: Let's try problem 34b on page 200 of your text: ln y = 3 − 0.5 x

This does not have a log on both sides. I will write it in exponential form:

e 3 − 0.5 x = y
e3
=y
e 0.5 x
e3
¡ ¢ x = y.
e 0.5

Example 9: Let's try problem 38 on page 201 of your text.
¡x¢
(a)     f     3
= log 3 3
x

= log 3 x − log 3 3
= log 3 x − 1
= f (x) − 1

(b)
Horizontally stretching the graph by a factor of 3 is equivalent to shifting down one unit. The
f (x), which is y on the graph, has been decreased by 1.

¡     ¢
Example 10: If log 9 5 = x and log 9 4 = y, express log 9 12 1 in terms of x and y:
2

1 25 5 2
12    =  =√
2 2    4
2 log 9 5 = log 9 5 2
1                √
log 9 4 = log 9 4
2
52             1
log 9 √ = 2 log 9 5 − log 9 4
4             2
52       1
log 9 √ = 2 x − y.
4       2

Example 11: Solve log 5 (x + 24) + log 5 x = 2.

You need a log on the right side with a base of 5. The log would need to be equivalent to a power of
2.
Lesson 5

log 5 (x + 24) + log 5 x = log 5 25
log 5 x (x + 24) = log 5 25
x (x + 24) = 25
x 2 + 24 x − 25 = 0
(x + 25) (x − 1) = 0
x = − 25      x = 1

The value of x cannot be −25 because you can't take the log of a negative number or zero. This
leaves 1 as the solution. Let's check it:

log 5 (1 + 24) + log 5 1 = 2
log 5 25 + log 5 1 = 2
2 + 0 = 2.

Example 12: Let's try problem 50 on page 202 of your text:

log 2 (x + 5) + log 2 (x − 2) ≥ 3.

Convert 3 to log 2 8:

log 2 (x + 5) + log 2 (x − 2) ≥ log 2 8
log 2 (x + 5) (x − 2) ≥ log 2 8
(x + 5) (x − 2) ≥ 8
x 2 + 3 x − 10 ≥ 8
x 2 + 3 x − 18 ≥ 0
(x + 6) (x − 3) ≥ 0
− 6       − 3

But the value of −6 and less will not work. The solution is x ≥ 3.

Study Exercises
Complete odd-numbered problems 1–53 in the Written Exercises on pages 200–203 of your text.

Section 5-7: Exponential Equations; Changing Bases pp. 203–207

Now we will use our logarithms to help us find the value of an exponent. In Section 5-2, we found
the values of an exponent by rewriting the problem with like bases:
Lesson 5

√
25 x =
5
5

I5 2M = (5) 1 / 5
x

1
5 2 x =
5
1
x =           .
10

But not all problems can be rewritten in the same base. In these cases the log function will be used in
the problem and our trusty calculator will find the value of the logs.

Example 1: Solve 25 x = 2.

Place the log function on both sides of the equation:

log 25 x = log 2.

Use law number 4:

x log 25 = log 2.

Divide to get x alone:

log 2
x=           .
log 25

Your calculator has a log key, and it is set for base 10. Base 10 is also the base in this problem. Use
your calculator to find both log values:

.301
x=
1.398
x = .215.

To the nearest hundredth is .22.
√
Example 2: Solve 8 x =     4
5:
√
4
log 8 x = log 5
x log 8 = log (5) 1 / 4
x log 8 = .25 log 5
.25 log 5
x =
log 8
(.25) (.699)
x =
.903
.17475
x =
.903
x = .19.

Example 3: Let's try problem 14 on page 205 of your text.

Use P (t) = P 0 e r t. P 0 is the original amount, when it is tripled it will look like 3 P 0.
Lesson 5

3 P 0 = P 0 e (.07) (t)

Divide P 0 over and it's gone:

3 = e (.07) (t).

Use the natural log:

ln 3 = ln e (.07) (t)
ln 3 = .07 t ln e
ln 3
=  t
.07 ln e
1.099
=t
.07
15.7 = t.

It will take 15.7 years or 15 years and 8.4 months to triple the investment.

As I mentioned earlier, your calculator is set up for base 10. But that won't stop you from being able
to solve problems in other bases. There is a change-of-base formula:

log c
log b c =         .
log b

Example 4: Solve 5 x = 98.

Rewrite in log form:

log 5 98 = x
log 98
x=
log 5

Use the change-of-base formula:

1.99
x=
.699
x = 2.85.

Example 5: Let's try problem 24 on page 206 of your text:

xx = π
log x x = log π
x log x = log π
log π
x=          .
log x
Lesson 5

log π
If you have a graphing calculator and you would like to graph the change to y = log x , here is the
procedure to use. Since y and x are to be the same also, graph y = x on the same coordinate axis.
Keep zooming in until you can trace and find the point of intersection to the nearest hundredth. It
will have a value of 1.85.

Example 6: Let's try problem 30 on page 206 of your text:

1
ex +      = 4.
ex

Multiply both sides by e x:

e2 x + 1 = 4 ex
e 2 x − 4 e x + 1 = 0.

Replace e x with y:

y 2 − 4 y + 1 = 0.

√
4±       [−4 (1) (1) + 16]
y=
2
√
4±       12
y=
2
√
4±2 3
y=
2
¡    √ ¢
2 2± 3
y=
2
y = I2 ± 3M.
√

Replace y with e x:
√
ex = 2 ±            3.

Rewrite in log form:
√
ln 2 ±          3 = x.

Example 7: Let's try problem 36 on page 207 of your text.

Replacing log a b and log b c with the change-of-base formula:

log b log c log c
×     =      .
log a log b log a

Rewrite as a single log using change-of-base formula: log a c.

Example 8: Let's try problem 40 on page 207 of your text.

Using the proven formula from problem 35:
Lesson 5

1                       1
= log 6 4  and          = log 6 9.
log 4 6                 log 9 6

Using law number 1:

log 6 4 + log 6 9 = log 6 36
log 6 36 = 2.

Study Exercises
Complete odd-numbered problems 1–43 in the Written Exercises section on pages 205–207 of your