LESSON EIII.E EXPONENTS AND LOGARITHMS

LESSON EIII.E – EXPONENTS AND LOGARITHMS

OVERVIEW

Exponential and logarithmic functions have many useful applications in fields ranging from investment banking to medicine. They are used to measure many things ranging from the strength of an earthquake to the noise level in a recording studio.

In this lesson, you will look at some applications as you review exponential and

logarithmic functions. You will start by graphing them. You will also review the properties of exponents and logarithms, and you will see how these properties can be used to solve exponential equations.

Here’s what you’ll learn in this lesson:

Exponential Functions

a. Graphing exponential functions b. Applications of exponential functions c. Solving some exponential equations Logarithmic Functions

a. Exponential and logarithmic form b. Graphing logarithmic functions c. Properties of logarithms Solving Equations

a. Using a calculator to approximate common and natural logarithms b. Change of base formula c. Solving logarithmic equations d. Solving exponential equations

EXPLAIN

EXPONENTIAL FUNCTIONS Summary

You have already graphed linear functions and quadratic functions. In this concept you will work with exponential functions. You will graph exponential functions, look at applications of exponential functions, and solve exponential equations.

Definition of an Exponential Function

Here are some examples of exponential functions:

f (x ) = 5^x g (x ) =

 

^{x h(x) = 7–x}

In general, an exponential function is a function of the form y = f (x ) = b^x.

Here, the constant b is called the base and is a positive number not equal to 1. The independent variable, x, is the exponent.

The domain of an exponential function is all real numbers.

The range of an exponential function is all positive real numbers.

The Graph of an Exponential Function

To graph an exponential function you can make a table of points, plot the points, and join them with a smooth curve, as you have done for other functions.

Here’s a table of points for the exponential function y = 3^x. The graph is shown in Figure EIII.E.1.

x y = 3^x

3 27

2 9

1 3

0 1

–1 –2 –3

From the graph of the exponential function y = 3^x, you can see that as x increases, the graph rises rapidly. As x becomes more negative, the graph gets closer to the x -axis, but never becomes zero or negative. The y -intercept is the point (0, 1).

1 27

1 9

1 3

1

3 Notice that in an exponential function,

the variable is the exponent. The function y = x³is not an exponential function because the variable, x, is the base.

The variable, x, does not have to be a whole number. For example, when x = , f (x) = 3 = 3. You can use your calculator to approximate

3≈1.73.

Remember, 3^–2= =

 

^{2= .}

Figure EIII.E.1

1 2 3

4 5 6

3 2 1 –1 –2

–3 x

y

y = 3^x

1 9

1 3

1 3²

1

1 2

2

Exponential Growth and Decay

Here’s a table of points for the exponential function y = f (x ) = 4^x. The graph of y = 4^xand the graph of y = 3^xare shown together in Figure EIII.E.2.

x y = 4^x

2 16

1 4

2

0 1

–1 –2

Notice how the graph of y = 4^xcompares to the graph of y = 3^x:

• Both graphs have y -intercept (0, 1).

• For positive values of x, the graph of y = 4^x rises more steeply than the graph of y = 3^x.

• For negative values of x, the graph of y = 4^x is closer to the x -axis than the graph of y = 3^x.

In general, for an exponential function y = b^x when b > 1, as you increase the value of b the graph rises more steeply for positive values of x and is closer to the x -axis for negative values of x .

For an exponential function y = b^x when b > 1, as you decrease the value of b the graph rises less steeply for positive values of x and is further from the x -axis for negative values of x .

An exponential function y = b^x with base b > 1 represents exponential growth.

For an exponential function y = b^x when b is between 0 and 1, the behavior of the graph is different. For example, here is a table of points for y = f (x ) =

 

^x. The graph is shown in Figure EIII.E.3.

–x y =

 

^x

–3 27

–2 9

–1 3

–0 1

–1 –2

–3 ¹ 27

1 9

1 3

1 3

1 3

1 16

1 4

1 2

Figure EIII.E.2

You can also write y =

 

^{xas y = 3–x.}

Figure EIII.E.3

1 2 3

4 5 6

3 2 1 –1 –2

–3 x

y

y =



^_31



^x

1 3

1 2 3

4 5 6

3 2 1 –1 –2

–3 x

y

y = 3^x y = 4^x

If you compare this table with the table for y = 3^x, you see the same y- values, but the corresponding x- values have changed sign. The graph of y =

 

^x is the reflection of the graph of y = 3^xabout the y -axis. You can see that the graph of

y =

 

^xdecreases as you move from left to right.

Figure EIII.E.4 compares the graph of y = 3^x with the graph of y =

 

^x.

An exponential function y = b^x with base 0 < b < 1 represents exponential decay.

Applications of Exponential Functions

There are two very common applications of an exponential function.

The first involves exponential growth as it has to do with the calculation of compound interest.

The function A (t ) = P (1 + r )^tis an exponential function describing the total amount of money in an account after t years. The constant P describes the amount of money you initially deposit, and the constant r, written as a decimal, describes the annual interest rate.

For example, suppose you deposit $500 in a savings account with an annual interest rate of 4%, compounded once a year. Here, the initial amount you deposit, P, is $500. The annual interest rate, 4%, expressed as a decimal is r = .04.

Then the total amount of money in the account after 9 years is given by:

A (9) = 500(1 + .04)⁹

= 500(1.04)⁹

≈500(1.4233)

= 711.65

So the total amount in the account after nine years is approximately $711.65.

The second application involves exponential decay and has to do with how a radioactive substance decreases in radioactivity over time. The function A (t ) = Pb^{– r t} is an exponential function describing the amount of radioactivity left in a substance after t years. The constant P describes the amount of radioactivity the substance started with.

The base, b , is a constant that depends on the radioactive chemical you are studying. So does r , which is called the decay constant.

The Base e

There is a base that is especially useful in applications such as radioactive decay. It is also important in calculus. This base is the irrational number e. The number e also arises naturally in many applications in science. The number e is an irrational number that lies between 2 and 3 and is approximately 2.718.

1 3

1 3

1 3

Figure EIII.E.4

Here you can use a calculator to determine that (1.04)⁹ 1.4233.

The decimal representation of e, 2.718..., like the decimal representation of the number πor the decimal

representation of the number 2, does not repeat nor does it stop.

1 2 3

4 5 6

3 2 1 –1 –2

–3 x

y

y =



^_31



^{x y = 3x}

The graph of y = e^x is the same shape as the graph of y = 2^xand the graph of y = 3^x. Because the base, e, is between 2 and 3, the graph of y = e^x lies between the graphs of y = 2^x and y = 3^x, as shown in Figure EIII.E.5.

You can use a calculator with an e^x key to obtain approximate answers to calculations involving e. For example, to calculate 200e^{– 0.7}:

1. Enter the exponent without its sign. 0.7

2. Press the ± key. –0.7

3. Press the e^xkey. 0.4965853

4. Multiply by 200. 99.317060

So 200e^{– 0.7}is approximately 99.32.

Continuous Compound Interest

Recall the formula A (t ) = P (1 + r )^t, which is used when interest is compounded annually.

There is a formula that can be obtained from this formula, and which represents continuous compounding (compounding as frequently as possible) of interest. This formula involves e and is given by the following:

A (t ) = Pe^{r t}

As before, P is the original deposit, r is the annual interest rate expressed as a decimal, and t is the number of years.

So to calculate the effect of continuous compounding on a deposit of $500 at 4% interest for 9 years you get:

A = 500e^0.04(9)

= 500e^0.36

≈500(1.433329)

= 716.66

So the total amount with continuous compounding is approximately $716.66.

Notice that this is a larger amount than the $711.66 that you got by compounding only once each year.

Exponential Equations

You have seen how to solve equations such as 2x = 4x – 7 and x²– 2x + 7 = 0.

Now, you will solve exponential equations.

An exponential equation is an equation that contains the variable in an exponent. Here are some examples:

7x 3 t – 2 2t + 1

Your calculator may work differently.

Figure EIII.E.5

Here you can use a calculator to determine that e^0.36≈^1.433329.

1 2 3

1 2 3

–1 –2 –3 –1 –2

–3 x

y

y = 2 x

y = e x

y = 3 x

One way to solve an exponential equation is to make use of the following property of exponents:

If b^x = b^y, then x = y.

(Here, b, x, and y are real numbers, b > 0, and b ≠1.)

You will also use some of the other properties of exponents that you have learned.

Here are steps you can use to solve some exponential equations:

1. Write both sides of the equation using the same base: b^x = b^y. 2. Set the exponents equal to each other: x = y .

3. Finish solving.

For example, to solve 4^{2 t – 1}= 8^{3 t}for t :

1. Write both sides of the equation

(

2²

)

^{2t – 1}=

(

2³

)

^3t

using the same base, 2. 2^{2 (2t – 1)}= 2^{3(3t )} 2^{4t – 2}= 2^9t 2. Set the exponents equal to each other. 4t – 2 = 9t

3. Finish solving for t . – 2 = 5t

t = – So t = – .

Sample Problems

1. Here is the graph of the exponential function f (x ) = 5^x. On the same set of axes graph:

a. y = 7^x b. y = 2^x c. y = 5^–x

■

■

✓

a. To graph y = 7^x, notice that the base 7 is greater than the base 5. So the graph rises more steeply for positive x, and is closer to the x -axis for negative x.

x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x y = 7^x

x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x

2 5

2 5

Some exponential equations cannot be solved by using this technique. For example, in the equation 3^x= 7, the number 7 cannot easily be written using the base 3. In the next section, you will solve this type of equation by using logarithms.

Answers to Sample Problems

■

■b. To graph y = 2^x, compare the base 2 to the base 5.

■

■ c. To graph y = 5^{– x}, notice that 5^{– x}=

 

^x.

2. The amount, A (t ), of radioactivity remaining in a radioactive substance after t years is given in kilograms by A (t ) = 3000e^–.0023t. Find the starting amount of radioactivity and the amount remaining after 500 years.

■

■

✓

a. To find the starting amount in kilograms, A (0) = 3000e^–.0023(0) substitute t = 0 into the formula for A.

■

■

✓

b. Simplify. = 3000e⁰

= 3000(1)

= 3000

■

■ c. To find the amount in kilograms A (500) = ___________

remaining after 500 years,

substitute t = 500 into the formula for A.

■

■ d. Simplify. Round your answer to two decimal ≈___________

places at the end of your calculations.

3. Solve this exponential equation for x : 9^{4x – 3}= 27^{4x + 1}

■

■

✓

a. Write both sides of the

(

3²

)

^{4x – 3}=

(

3³

)

^{4x + 1}

equation using the base 3.

■

■ b. Simplify the exponents. __________ = __________

■

■ c. Set the exponents equal to each other. __________ = __________

■

■ d. Finish solving for x. x = _________

1 5

x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x

Answers to Sample Problems b.

c.

c. 3000e–.0023(500)

d. 949.91

b. 3^{8x – 6}, 3^{12x + 3} c. 8x – 6 , 12x + 3 d. – _⁹

4

x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x y = 5^–x

x y

–1 4 5 6

3 2 1

1 –2

–3 2 3

y = 5^x

y = 2^x

LOGARITHMIC FUNCTIONS Summary

In this concept you will look at the inverse function of an exponential function, which is called a logarithmic function. You will graph logarithmic functions and study their properties.

Definition of a Logarithm

You have already used the inverse of a function to write the same information in two different ways.

For example, you can use the “squaring function” to write:

3²= 9

You can use the “square root function” to write the same information as:

9 = 3

So the “squaring function” is the inverse of the “square root” function.

Similarly, the logarithmic function, y = log_bx, is the inverse of the exponential function.

Here are some examples of equations written in the exponential form and in the corresponding logarithmic form:

Exponential Form Logarithmic Form

3²= 9 log₃ 9 = 2

10⁵= 100000 log₁₀100000 = 5

2⁴= 16 log₂16 = 4

(25) = 5 log₂₅5 =

In general, this exponential statement: b^L= x is equivalent to this logarithmic statement: log_bx = L

Any logarithmic statement can be rewritten as an exponential statement and vice versa.

You can use this idea to calculate some logarithms.

For example, to find log₂ :

Call this expression x. x = log₂

Rewrite the statement in exponential form. 2^x = Solve the exponential equation.

• Write both sides using the base 2. 2^x = 2^{– 4}

• Set the exponents equal to each other. x = – 4 So x = log_2¹ = – 4.

16

1 16

1 16

1 16

1 2

12

exponent 2³= 8

base

argument logarithm log₂8 = 3

base

Some logarithms cannot be evaluated this way. In the next section, you will approximate them using your calculator.

The Graph of a Logarithmic Function

When you switch between exponential and logarithmic form you see that the exponential function and the logarithmic function are inverse functions.

Here are the steps to confirm that these functions are inverses of each other.

Start with the exponential function . f (x ) = b^x

Replace f (x ) with y. y = b^x

Switch y and x. x = b^y

Solve for y . log_b x = y

Replace y with f^–1( x ). f^–1( x ) = log_b x

Since the exponential function and the logarithmic function are inverses of each other, you can use this to your advantage in graphing y = log_bx . Recall that:

• The graph of the inverse function y = f^–1(x ) is obtained by reflecting the graph of y = f (x ) about the line y = x.

Here’s how to graph y = log₃x.

Start with a table of values for the function y = 3^x. –x y = 3^x –2 9 –1 3 –0 1 –1

–2

The graph is shown in Figure EIII.E.6.

To graph the inverse function, y = log₃x , reflect the graph of y = 3^xabout the line y = x.

This is shown in Figure EIII.E.7.

From the graph, you can see that as x increases, the graph of y = log₃x rises slowly. As x gets closer to zero, the graph of y = log₃x becomes more and more negative.

The x -intercept is the point (1, 0).

The domain of the logarithm function is the positive real numbers.

The range of the logarithm function is the real numbers.

In general, if b > 1, the graph of the function y = f (x ) = log_bx behaves like the above example, y = log₃x.

1 9

1 3 Remember, to find the inverse, f^–1, of f:

1. Replace f (x) with y.

2. Switch y and x.

3. Solve for y.

4. Replace y with f^–1(x).

Figure EIII.E.6

Figure EIII.E.7

2 4 6

2 4 6

–2 –4 –6 –4 –2

–6 x

y y = 3^x

y = log₃ x y = x

2 4 6

2 4 6

–2 –4 –6 –2 –4

–6 x

y

y = 3^x

Here’s another example. To graph y = log x : Start with a table of values for the function y =

 

^x.

–x y =

 

^x

–2 –1

–0 1

–1 3

–2 9

The graph is shown in Figure EIII.E.8.

To graph the inverse function, y = log x , reflect the graph of y =

 

^{xabout the}

line y = x.

This is shown in Figure EIII.E.9.

From the graph, you can see that as x increases, the graph of y = log x falls slowly.

As x gets closer to zero, the graph of y = log x becomes more and more positive.

The x -intercept is the point (1, 0).

The domain of the logarithm function is the positive real numbers.

The range of the logarithm function is the real numbers.

In general, if 0 < b < 1, the graph of the function y = f (x ) = log_b x behaves like the above example, y = log x.

Now you have seen the shape of the graph of a logarithmic function when the base, b, is greater than 1, or when the base, b, is between 0 and 1. You can use this information to graph a logarithmic function directly by plotting a few points that satisfy the function and joining these points with a smooth curve.

Base 10 and Base e

For most applications of logarithms you will use logarithms whose base is either 10 or e.

Logarithms to the base 10 are called common logarithms and are usually written without the base 10, simply as log x rather than log₁₀ x.

Logarithms to the base e are called natural logarithms and are usually written ln x rather than log_e x.

1 3

13

1 3

1

1 3

3

1 3

1 9

1 3

1 3

1 3

Figure EIII.E.8

Figure EIII.E.9

2 4 6

2 4 6

–2 –4 –6 –2 –4

–6 x

y

y =



^_31



^x

y = x

y = log_{_1} x

3

2 4 6

2 4 6

–2 –4 –6 –2 –4

–6 x

y

y =



^{1 _3}



^x

Properties of Logarithms

There are several algebraic properties of logarithms that are true for any base b. They correspond to the familiar properties of exponents.

Property Examples

log_b b = 1 log₇7 = 1

log_ee = 1

log_b1 = 0 log₁₀1 = 0

ln 1 = 0

b^{logb x} = x 5^{log 5 22}= 22

e^{ln 8}= 8

log_bbⁿ = n log₁₄14²⁹= 29

ln e⁶= 6 Log of a Product

log_bu v = log_bu + log_bv log₁₁17x = log₁₁ 17 + log₁₁x Log of a Quotient

log_b = log_b u – log_bv log₁₁ = log₁₁ 17 – log₁₁x Log of a Power

log_buⁿ= n log_bu log₈13²²= 22 log₈13 log_b = log_b v^–1= – log_bv log₆ = – log₆32 Here’s an example that uses the first four properties of logarithms to simplify an expression containing logarithms.

To find the value of the expression ln e + log₈1 – 7^{log7 11}:

• Rewrite In e as log_ee and use the property log_b b = 1. = 1 + log₈1 – 7^{log7 11}

• Use the property log_b1 = 0. = 1 + 0 – 7^{log7 11}

• Use the property b^{logb x}= x. = 1 + 0 – 11

= – 10 So ln e + log₈1 – 7^{log7 11}= –10.

Here’s another example.

To find the value of the expression ln 1 – log₁₁11⁵:

• Rewrite In 1 as log_e1 and use the property log_b1 = 0. = 0 – log₁₁11⁵

• Use the property log_bbⁿ= n. = 0 – 5

= –5 So, ln 1 – log₁₁11⁵= –5.

1 32

1 v

17 x

u v

Now here’s an example that uses the product and quotient properties of logarithms.

To rewrite the expression log using several logarithms:

• Use the log of a quotient property. = log (2x – 1)(3x + 5) – log (6x + 7)

• Use the log of a product property. = log (2x – 1) + log (3x + 5) – log (6x + 7) So log = log (2x – 1) + log (3x + 5) – log (6x + 7).

Here’s another example.

To simplify ln

 

^:

• Use the log of a quotient property. = ln 1 – l n 23

• Use the property log_b1 = 0. = 0 – ln 23 So, ln

 

= – ln 23.

Here’s an example that uses the logarithm of a power property.

To simplify log₅³ 7 – log₅125:

• Write using exponents. = log₅7 – log₅5³

• Use the log of a power property. = log₅7 – 3log₅5

• Use the property log_b b = 1. = log₅7 – 3 1

= log₅7 – 3 So, log₅³ 7 – log₅125 = log₅7 – 3.

In general, you can simplify expressions containing logarithms using any combination of the logarithmic properties.

Here’s another example:

To use logarithmic properties to write the expression 3ln x + ln (x – 1) – 5 ln (2x + 3) as a single logarithm:

• Use the log of a power property. = ln x³+ ln (x – 1) – ln (2x + 3)⁵

• Use the log of a product property. = ln



^{x3(x – 1)}



– l n (2x + 3)⁵

• Use the log of a quotient property. = ln So, 3ln x + ln (x – 1) – 5 ln (2x + 3) = ln ^{x3(x – 1)}.

(2x + 3)⁵ x³(x – 1)

(2x + 3)⁵

1 3

1 3

1 3

1 3

13

1 23

1 23 (2x – 1)(3x + 5)

6x + 7

(2x – 1)(3x + 5)

6x + 7

The log of a product property states that the logarithm of a product is the sum of the separate logarithms. The log of a quotient property states that the logarithm of a quotient is the difference of the two separate logarithms.

Be careful when using the quotient property.

You can use it this way:

log

 

= log 7 – log 4

But a quotient of logs is not the same thing: ^{log 7}_ ≠log 7 – log 4

log 4

7 4

Sample Problems

1. Use the exponential form to find log₄

 

^.

■

■

✓

a. Call this expression x. x = log₄

 

■

■b. Rewrite in exponential form. ________ = ________

■

■c. Rewrite using the same ________ = ________

base on each side.

■

■d. Set the exponents equal ________ = ________

to each other.

■

■e. Solve for x. x= ___________

2. Graph the function y = log x .

■

■

✓

a. Complete the table for the x y =

 

^x

exponential function –2 16

y =

 

^{x. –1 4}

0 1 1

■

■ b. Plot the points and join them with a smooth curve to graph y =

 

^x.

■

■ c. Draw the line y = x . Reflect the graph of y =

 

^xabout

the line y = x to graph y = log_¹ x.

4

1 4

x y

–5 5 10 15

–5 –10 –15

5 –10

–15 10 15

1 4

x y

–5 5 10 15

–5 –10 –15

5 –10

–15 10 15

1 4

1 4

1 4

14

1 8

1 8

Answers to Sample Problems

b. 4^x, c. 2^2x, 2^{– 3}

d. 2x , –3

e. –

b.

c.

x y

–5 5 10 15

–5 –10 –15

5 –10

–15 10 15

y = ^{— 14}^x

y = x

y = log x1_ 4

x y

–5 5 10 15

–5 –10 –15

5 –10

–15 10 15

y =  ^1_4^x

3 2

1 8

3. Use properties of logarithms to show that:

ln = 5 + 2 ln (x + 1) – ln (x – 3) – ln (x – 2).

■

■

✓

a. Use the log of a quotient l n

property on the left side. = ln e⁵(x + 1)² – ln (x²– 5x + 6)

■

■ b. Use the log of a product property on the first

logarithm. = _________________________

■

■ c. Factor x²– 5x + 6 in the third logarithm.

= _________________________

■

■ d. Use the log of a product property on the third

logarithm. = _________________________

■

■ e. Use the property log_bb^x = x on the first logarithm.

= _________________________

■

■

✓

f. Use the log of a power

property on the second term. = 5 + 2 ln (x + 1) – ln (x – 3) – ln(x – 2)

e⁵(x + 1)²

x²– 5x + 6 e⁵(x + 1)²

x²– 5x + 6

Answers to Sample Problems

b. ln e⁵+ ln (x + 1)²– ln (x²– 5x + 6)

c. ln e⁵+ ln (x + 1)²– ln [(x – 3)(x – 2)]

d. ln e⁵+ ln (x + 1)² – ln (x – 3) – ln (x – 2)

e. 5 + ln (x + 1)²– ln (x – 3) – ln(x – 2)

SOLVING EQUATIONS Summary

In this concept you will use a calculator to approximate certain logarithms and exponents, and you will change the base of logarithms. Then you will solve a variety of equations that contain logarithms or exponents.

Calculating Common Logarithms and Exponents

You have already seen how to calculate some logarithms by switching to exponential form.

For example to find log 100:

• Call this expression x . x = log 100

• Rewrite in exponential form. 10^x = 100

• Write both sides using base 10. 10^x = 10²

• Set the exponents equal to each other. x = 2 So log 100 = 2.

This method was successful because 100 could be written as an integer power of the base, 10. That is, 100 = 10².

This method cannot be used to find log₁₀70 because 70 is not an integer power of 10.

That is, there is no integer n, such that 10ⁿ= 70.

You can approximate the common logarithm of a number on your calculator.

Here are the steps you can use on many calculators.

1. Enter the number.

2. Press the log key.

For example, to find the common logarithm of the number 70:

1. Enter the number. 70

2. Press the log key. 1.84509

So, log₁₀70 ≈1.85.

You can also reverse this process to find a number if you know its common logarithm.

Recall that the common logarithm, log₁₀x , is often written log x.

To find a number, given its common logarithm:

1. Enter the logarithm in your calculator, using the ±key if necessary.

2. Press the 10^x key.

For example, here’s how to find x if you know that log₁₀ x = 1.63:

1. Enter the logarithm, using 1.63 the ± key if necessary.

2. Press the 10^x key. 42.657951

So, if log₁₀ x = 1.63, then x ≈42.66.

Calculating Natural Logarithms and Exponents

To calculate a natural logarithm you can use the same general steps as you did for common logarithms and exponents, but use the ln key instead of the log key, and use the e^x key instead of the 10^x key.

For example, to find ln 43.7:

1. Enter the number. 43.7

2. Press the ln key. 3.777348

So, ln 43.7 ≈3.78.

Now here’s how to find x if you know that ln x = –1.95:

1. Enter the logarithm. –1.95

2. Press the ± key. –1.95

3. Press the e^x key. –0.14227407

So, if ln x = – 1.95, then x ≈0.14.

The Change of Base Formula

Most calculators have a key for common logarithms (log) and a key for natural logarithms (ln). In order to approximate a logarithm to any other base, you can use the following change of base formula:

log_bx =

(Here, b, c, and x are positive numbers, b≠^{1, and c} ≠^1.)

This formula lets you find a logarithm to the base b by choosing any other convenient base, c. Usually you will choose base 10 or base e for c .

log_cx

log_cb

The functions log₁₀ x and 10^x are inverses of each other. On many calculators, they appear on the same key.

Recall that the natural logarithm, log_ex, is often written ln x.

The functions ln x and e^xare inverses of each other. On many calculators, they appear on the same key.

Here’s how to approximate log₃8.93 by using common logarithms and the change of base formula:

1. In the change of base formula, log₃8.93 = let c = 10, b = 3, and x = 8.93.

2. Find log 8.93. log₁₀8.93 ≈0.9508514

3. Find log 3. log₁₀3 ≈0.4771212

4. Divide. log₃8.93 =

≈1.992892 So, log₃8.93 ≈1.99.

Now here’s how to approximate log₃8.93 by using natural logarithms and the change of base formula:

1. In the change of base formula, log₃8.93 = let c = e, b = 3, and x = 8.93.

2. Find ln 8.93. ln 8.93 ≈2.189416

3. Find In 3. ln 3 ≈1.098612

4. Divide. log₃8.93 =

≈1.992892 So, log₃8.93 ≈1.99.

Equations that Contain One Logarithm

Here are some examples of equations that contain one logarithmic term:

3 – 5 log₄x = 7 ln x = 5 2.2 + log_x 7 = 4.2

Here are some steps that help to solve equations that contain one logarithmic term:

1. Isolate the logarithm on one side of the equation.

2. Rewrite the equation in exponential form.

For example, to solve the equation 3 – 5 log₄x = 7:

1. Isolate the logarithm on one side of the equation. – 4 = 5 log₄x log₄x = –

2. Rewrite the equation in exponential form. x = 4^–

3. Approximate using a calculator. = 4^–.8

≈0.3298769 So, x ≈0.33.

4 5

4 5 ln 8.93

ln 3 ln 8.93

ln 3 log 8.93

log 3 log₁₀8.93

log₁₀3

Notice that you get the same final answer whether you use common logarithms or natural logarithms in the change of base formula.

Now here’s an example where the variable, x , is the base.

To solve the equation 2.2 + log_x 7 = 4.2:

1. Isolate the logarithm on one side of the equation. log_x 7 = 2 2. Rewrite the equation in exponential form. 7 = x²

3. Solve for x. Discard any x = ⁷

negative values. = 2.6457513

So, x ≈ 2.65.

Equations that Contain More Than One Logarithm

To solve equations that contain more than one logarithm you can often use properties of logarithms to combine several logarithms into a single logarithm.

Here are some of the useful properties of logarithms.

Property Example

Log of a Product

log_buv = log_bu + log_bv log₁₁17x = log₁₁17 + log₁₁x Log of a Quotient

log_b = log_b u – log_bv log₁₁ = log₁₁17 – log₁₁x Log of a Power

log_b uⁿ= n log_bu log₈13²²= 22 log₈13

And here are two additional properties that you can use.

Property Example

If log_bu = log_bv If ln 3x = ln 5.7

then u = v then 3x = 5.7

If b^x = b^y If 10^{x + 1}= 10^{3x – 2}

then x = y then x + 1 = 3x – 2

Here are some steps to solve equations that contain several logarithms.

1. Rewrite the equation with all the log terms on one side.

2. Use properties of logs to combine into a single log.

3. Rewrite the equation in exponential form.

4. Finish solving.

Here’s an example.

17 x

u v

Negative values of x are discarded since the base, x, must be positive.

To solve 1 – log₂(x + 2) = log₂(6 – 3x ):

1. Rewrite the equation with log₂(6 – 3x ) + log₂(x + 2) = 1 all the log terms on one side.

2. Use properties of logs to log₂

[

(6 – 3x )(x + 2)

]

= 1 combine into a single log.

3. Rewrite the equation in exponential form. (6 – 3x )(x + 2) = 2¹

4. Finish solving. 6x + 12 – 3x²– 6x = 2

3x²= 10 x²=

x = ±

^≈±1.83

To ensure that the values of x you have found are actually solutions to the original equation, you need to check them. The definition of log_bx requires that x > 0. Since the original equation contains the terms log₂(x + 2) and log₂(6 – 3x ), you must check that x + 2 > 0 and that 6 – 3x > 0.

Check x = 1.83: Check x = –1.83:

Is x + 2 > 0? Is x + 2 > 0?

Is 1.83 + 2 > 0? Is –1.83 + 2 > 0?

Is 3.83 > 0? Yes. Is 0.17 > 0? Yes.

Is 6 – 3x > 0? Is 6 – 3x > 0?

Is 6 – 3(1.83) > 0? Is 6 – 3(–1.83) > 0?

Is .51 > 0? Yes. Is 11.49 > 0? Yes.

So x≈1.83 and x ≈–1.83 are both solutions.

If all the terms in an equation are logarithmic terms, then you can try these steps to solve the equation:

1. Combine the logs into a single log on each side.

2. Use the property: if log_bu = log_b v then u = v.

3. Finish solving.

Here’s an example.

To solve ln 7 + ln 2x = ln (6 – 3x ):

1. Combine the logs into a single log on each side. ln 14x = ln (6 – 3x ) 2. Use the property: if log_bu = log_bv then u = v. 14x = 6 – 3x

3. Finish solving. 17x = 6

x = _⁶

17

10 3

10 3

Once again you need to check that this value of x is a solution. Since the original equation contains the terms ln 2x and ln (6 – 3x ) you must check that 2x > 0 and 6 – 3x > 0.

Check x = : Is 2x > 0?

Is 2

 

^{> 0?}

Is > 0? Yes.

Is 6 – 3x > 0?

Is 6 – 3

 

^{> 0?}

Is > 0? Yes.

So, x = is a solution.

Equations with a Variable in the Exponent

You can use logarithmic and exponential properties to solve equations that contain a variable in an exponent. Here are two examples of such exponential equations.

4^x = 3^{2 x – 1} 5e^{4x + 3}– 8 = 2

Here’s a way to solve 4^x = 3^{2 x – 1}:

1. Take the log of both sides. log 4^x = log 3^{2x – 1} 2. Use the logarithm of a power property. x  log 4 = (2x – 1)  log 3 3. Finish solving for x.

• Distribute on the right. x · log 4 = 2x · log 3 – log 3

• Collect the x -terms on one side. x · log 4 – 2x · log 3 = – log 3

• Factor out x . x · (log 4 – 2 log 3) = – log 3

• Solve for x . x =

You can approximate this answer using your calculator:x ≈ ≈1.35.

Here’s a way to solve 5e^{4x + 3}– 8 = 2:

1. Isolate the term with the exponent. 5e^{4x + 3}= 10 e^{4x + 3}= 2 2. Take the natural log of both sides. ln e^{4x + 3}= ln 2 3. Use the property log_bbⁿ= n. 4x + 3 = ln 2

4. Solve for x. x =

You can approximate this using your calculator: x≈ 0.693147 – 3 ≈– 0.58.

4

ln 2 – 3

4

– 0.47712

0.602059 – 2(0.47712) – log 3

log 4 – 2 log 3

6 17

84 17

6 17

12 17

6 17

6 17

Here, you cannot easily write each side of the equation with the same base. So that’s why you use these steps.

In step 1 you can use any base for your logarithm. But you will usually choose base 10 or base e because they are available on your calculator.

Sample Problems

1. Use your calculator to approximate: log₃15.7

■

■

✓

a. In the change of base formula, log₃15.7=

let c = e, b = 3 and x = 15.7.

■

■ b. Use your calculator to find ln 15.7. ln 15.7 = _______

■

■ c. Use your calculator to find ln 3. ln 3 = _______

■

■ d. Calculate log₃15.7 =

≈ _______

2. Solve for x : log₃x = 1 – log₃(x – 1)

■

■

✓

a. Move the log terms to the log₃x + log₃(x – 1) = 1 left side of the equation.

■

■ b. Use the logarithm of a ______________ = 1 product property to simplify

the left side of the equation.

■

■ c. Write in exponent form. ________ = ________

■

■ d. Write the quadratic __________________

equation in standard form.

■

■ e. Solve the equation by using the quadratic formula.

Use your calculator to approximate x .

x ≈ ________ or x ≈ ________

■

■ f. Check the solutions.

3. Solve for x : 4 + 3e^{4x – 2}= 9

■

■

✓

a. Isolate the term with 3e^{4x – 2}= 5 the exponent.

■

■ b. Divide both sides by 3. _________ = _________

■

■ c. Take the natural logarithm _________ = _________

of both sides.

■

■ d. Finish solving for x. x = _________

■

■ e. Use your calculator to x ≈ _________

approximate x.

ln 15.7

ln 3 ln 15.7

ln 3.

ln 15.7

ln 3

Answers to Sample Problems

b. 2.75366 c. 1.098612

d. 2.51

b. log₃[x ( x – 1)]

c. x (x – 1), 3¹ d. x²– x – 3 = 0

e. 2.3 –1.3

f. x and x – 1 must be positive.

So x≈2.3 is the only solution.

b. e^{4 x – 2}, c. In e^{4 x – 2}, In

d.

e. .63 2 + ln 5

3

4

5 3

5 3

Explain Exponential Functions

1. Complete the following table for the function y =

 

^x.

–x y –2

–1

1

2. The formula A = 200e^{– t} gives the amount of radioactivity (in milligrams) remaining in a substance after t days in a laboratory experiment. Find approximately how much is left after 5 days.

3. Solve for t : 3^2t = 81

4. The graphs of y = 4^x, y = 8^x and y = 6^–xare shown in Figure EIII.E.10. Identify which graph represents each function.

Figure EIII.E.10

5. The formula A = 2000(1 + .053)¹⁰gives the total amount in a savings account after $2000 has compounded for 10 years at a 5.3% annual interest rate. Use your calculator to find A.

6. Solve for x :

 

^{3x = 32}

7. In each statement, circle the correct choice.

a. For x < 0, the graph of y = 5^xis closer to/further from the x-axis than the graph of y = 11^x.

b. For x > 0 the graph of y = 5^x rises less steeply/more steeply than the graph of y = 11^x.

8. The formula A = P e^{r t} gives the total amount in a savings account after a deposit of P dollars compounds continuously for t years at an annual interest rate r . If you deposit $420, which is compounded continuously at an interest rate of 4.7%, how much will you have after 20 years?

9. Solve for t : 4^2t = 8^{t + 1}

10. Graph the function f (x ) = 2e^xby using your calculator to find the points on the graph for the values x = –1, 0, 1 and 2. Plot the points on the graph, and then join these points with a smooth curve.

11. The number of printed circuit boards, N, that can be tested in one day by an assembly line worker who has x days of experience is given by this formula:

N = 200 – 15e^–.01x

Find the number of circuit boards that can be tested by a worker with 5 days of experience.

12. Solve for t : 9^{2 t – 1}= 27^{5t + 2}

1 2

1 2 3

4 5 6

3 2 1 –1 –2

–3 x

y

A

C B

1 4

1 4

HOMEWORK

Homework Problems

Circle the homework problems assigned to you by the computer, then complete them below.

Logarithmic Functions

13. Write this exponential statement in logarithmic form: 3^{– x} = 17 14. Write this logarithmic statement in exponential form:

log u = 5 15. Simplify:

a. log₇7¹³ b. 8^{log8 2x}

16. The graph of a function y = b^x is shown in Figure EIII.E.11.

Graph the function y = log_b x on the same grid.

Figure EIII.E.11

17. Below is a table of values for an exponential function y = b^x. Use this table to graph the inverse function y = log_b x .

–x y

–0.7 0.88

–0.4 1.08 –1.3 1.27 –2.8 1.67

18. Use properties of logarithms to rewrite this expression using two logarithms:

log₄(x + 6)(x – 7)

19. Find log₁₀10000.

20. Graph y = log₂x .

21. Use properties of logarithms to rewrite this expression as a single logarithm:

ln x – ln ( x + 1) + ln ( x – 2) – ln ( x + 3) 22. Find log₉

 

^.

23. Graph y = log x .

24. Use properties of logarithms to rewrite using two logarithms:

log₃ + log₄64

Solving Equations

25. Use your calculator to approximate: log₁₀37.4 26. Solve for x : 4 + 3 log₂x = 7

27. Solve for x : 3^x =

28. Use your calculator to approximate: ln 8.3 + ln 4.1 29. Solve for x : log₅(7x ) – 2 = 1

30. Solve for x : 2 · 5^x = 250

31. Use your calculator to approximate: 7e^{– 6.2} 32. Solve for x : log x + log (x + 3) = 1 33. Solve for x : 3 + e^{5x – 1}= 9

34. Use your calculator to approximate: log₇85.9 35. Solve for x : ln x + ln (x + 1) = 0

36. Solve for x : 5 · 6^{2x – 3}+ 1 = 8

1 9 9(2x – 5)⁷

(3x – 1)⁴

13

1 27

x y

2 3

Exponential Functions

1. Complete the table of values for the function y = f (x ) =

 

^x.

x y =

 

^x

–3 –2 –1 0 1 2

2. Use your calculator to complete the table of approximate values for the function y = f (x ) = 2e^–x.

x y = 2e^–x –2

–1 0 2

3. The graphs of y = 7x and y = 2x are shown below. Identify which graph is which.

4. Graph the exponential function y = f (x ) = 2 – x.

5. The compound interest formula A P (1  r )^tgives the total amount of money, A, in a retirement savings account after a deposit of P dollars has compunded annually for t years at an interest rate r (where r is expressed as a decimal). Find the total amount of money, A, in your account if your initial deposit of $1000 has compounded annually at an 8.3% annual interest for 30 years.

6. The approximate number of bacteria, N, in a sample of contaminated water is given by the formula

N = 5000 + 2000e^.5twhere t is the number of hours since the sample was collected. Find the number, N, of bacteria that are present 12 hours after a sample of contaminated water is collected.

7. The function A P



^{1 +}



^ntgives the approximate total amount of money, A, in a savings account where the interest is compounded n times each year. P represents the initial deposit, r the annual interest rate (expressed as a decimal), and t the number of years. Use this formula to find the total amount, A, obtained from a deposit of $2000 left to compound 3 times a year for 30 years at a 4.8% annual interest rate.

8. Solve for v : 8^{2v – 5} 4^{v 3} 9. Solve for x : e^{2x 1}= 10. Solve for t : 4^{2 – 3t}

 

^{5t – 2}

Logarithmic Functions

11. Rewrite this statement in logarithmic form:

 

^{7= x}

12. Rewrite this statement in exponential form: log_nP = Q 13. Find log₈32.

3 5

1 8

1 e^{x – 1}

r n

4 8 12

4 8 12

–4 –8 –12 –8 –4

–12 x

y A B

2 3

2 3

APPLY

Practice Problems

Here are some additional practice problems for you to try.