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6. Exponential & Logarithmic Functions
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United States of AmericaTennessee
Algebra 2

6.03 Logarithms

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Concept summary

The inverse of an exponential is called a logarithm.

Logarithm

A value representing the exponent to which a fixed number (the base) must be raised to produce a given number.

If we have an exponential equation of the form: x=b^y This is equivalent to a logarithmic equation of the form: \log_b \left(x\right)=y This relationship leads to the following property, that we can use to change between forms:

This relationship means that a logarithm of the form \log_b \left(x\right) is equal to the exponent, n, to which we would raise the base, b, in order to obtain x.

A logarithmic function is a function that contains a variable expression inside a logarithm; a function of the form y=\log_b\left(x\right), where x is restricted to the domain of the positive real numbers and the base b is any positive real number not equal to 1.

Natural logarithms are logarithms with a base of the mathematical constant e. The natural logarithmic function y=\log_e x which is commonly written as y=\ln x.

Worked examples

Example 1

Write the inverse function for each of the following:

a

f\left(x\right)=\ln\left(\dfrac{x}{2}\right)

Approach

We want to replace f\left(x\right) with y, and then swap x and y to find the inverse. We can then solve for y to find the inverse function.

Solution

\displaystyle f\left(x\right)\displaystyle =\displaystyle \ln\left(\frac{x}{2}\right)State the equation
\displaystyle y\displaystyle =\displaystyle \ln\left(\frac{x}{2}\right)f\left(x\right)=y
\displaystyle x\displaystyle =\displaystyle \ln\left(\frac{y}{2}\right)Swap x and y
\displaystyle e^x\displaystyle =\displaystyle \frac{y}{2}\log_b x=n \iff x=b^n
\displaystyle 2e^x\displaystyle =\displaystyle yMultiply by 2
\displaystyle y\displaystyle =\displaystyle 2e^xSymmetric property of equality

The inverse function is f^{-1}\left(x\right)=2e^x.

b

f\left(x\right)=7^{3x+1}

Approach

We want to replace f\left(x\right) with y, and then swap x and y to find the inverse. We can then solve for y to find the inverse function.

Solution

\displaystyle f\left(x\right)\displaystyle =\displaystyle 7^{3x+1}State the equation.
\displaystyle y\displaystyle =\displaystyle 7^{3x+1}f\left(x\right)=y
\displaystyle x\displaystyle =\displaystyle 7^{3y+1}Swap x and y
\displaystyle \log_7 \left(x\right)\displaystyle =\displaystyle 3y+1\log_b x=n \iff x=b^n
\displaystyle \log_7 \left(x\right)-1\displaystyle =\displaystyle 3ySubtract 1
\displaystyle \frac{\log_7 \left(x\right)-1}{3}\displaystyle =\displaystyle yDivide by 3
\displaystyle y\displaystyle =\displaystyle \frac{\log_7 \left(x\right)-1}{3}Symmetric property of equality

The inverse function is f^{-1}\left(x\right)=\dfrac{\log_7 \left(x\right)-1}{3}.

Example 2

Consider the logarithmic equation: y=\log_\frac{1}{3} 9

a

Rewrite the equation in exponential form.

Approach

To rewrite this equation, we can use the property \log_b x=n \iff x=b^n .

Solution

We can rewrite this equation as 9=\left(\dfrac{1}{3}\right)^y

b

Find the value of y.

Approach

To solve for y, we want to find the exponent that we would raise \dfrac{1}{3} by, to obtain a result of 9. We can rewrite this equation as 9=\dfrac{1}{3^y}=3^{-y}. This means we want to find the exponent that we would raise 3 by, to obtain 9.

Solution

We can see that 9=3^2 which means y=-2.

\displaystyle 9\displaystyle =\displaystyle \left(\dfrac{1}{3}\right)^yFrom part (a)
\displaystyle 3^2\displaystyle =\displaystyle 3^{-y}Rewrite both sides as a power of 3
\displaystyle 2\displaystyle =\displaystyle -yIf a^m=a^n, then m=n.
\displaystyle -2\displaystyle =\displaystyle yDivide by -1

Example 3

For each of the following logarithmic equations, solve for x:

a

\log_7\left(3x+10\right)=2

Approach

To solve this equation for x we can use the identity \log_b x=n \iff x=b^n . We can then evaluate the result.

Solution

\displaystyle \log_7\left(3x+10\right)\displaystyle =\displaystyle 2State the equation
\displaystyle 3x+10\displaystyle =\displaystyle 7^2\log_b x=n \iff x=b^n
\displaystyle 3x+10\displaystyle =\displaystyle 49Evaluate the square
\displaystyle 3x\displaystyle =\displaystyle 39Subtract 10
\displaystyle x\displaystyle =\displaystyle 13Divide by 3
b

\log_{10}x+\log_{10}11=\log_{10}44

Approach

To solve this equation for x, we want to get it in the form \log_{10}\left(x\right)=c, so we can use \log_{10}\left(x\right)=c \iff x=10^c.

Solution

\displaystyle \log_{10}\left(x\right)+\log_{10}\left(11\right)\displaystyle =\displaystyle \log_{10}\left(44\right)State the equation
\displaystyle \log_{10}\left(x\right)\displaystyle =\displaystyle \log_{10}\left(44\right)- \log_{10}\left(11\right)Subtract \log_{10}11
\displaystyle x\displaystyle =\displaystyle 10^{\left(\log_{10}\left(44\right)- \log_{10}\left(11\right)\right)}Using \log_{10}\left(x\right)=c \iff x=10^c
\displaystyle x\displaystyle =\displaystyle 4Evaluate using a calculator

Outcomes

A2.F.LE.A.1.B

Understand that a logarithm is the solution to ab^(ct) = d, where a, b, c, and d are numbers.

A2.F.LE.A.1.C

Evaluate logarithms using technology.

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP5

Use appropriate tools strategically.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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