Logarithmic Functions as Inverses of Exponential Functions (College Board AP® Precalculus): Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Logarithmic functions as inverses of exponential functions

  • The logarithmic function  f(x)=logbx

    • and the exponential function g(x)=bx

      • where b>0 and b1

    • are inverse functions of each other

  • This means they "undo" each other:

    • logb(bx)=x

      • taking the log of an exponential with the same base returns the exponent

    • blogbx=x

      • raising a base to a logarithm with the same base returns the original value inside the logarithm

  • More formally, if  f(x)=logbx and g(x)=bx

    • then  f(g(x))=g(f(x))=x

How do the input-output behaviors of logarithmic and exponential functions compare?

  • Exponential and logarithmic functions handle change in opposite ways

  • In an exponential function

    • when input values increase by equal amounts (additively)

    • the output values change by equal ratios (multiplicatively)

  • In a logarithmic function, the relationship is reversed

    • when input values change by equal ratios (multiplicatively)

    • the output values increase by equal amounts (additively)

  • E.g. for g(x)=2x

    • increasing x by 1 (additive change)

    • always doubles the output (multiplicative change)

  • For  f(x)=log2x

    • doubling x (multiplicative change)

    • always increases the output by 1 (additive change)

x

g(x)=2x

x

 f(x)=log2x

0

1

1

0

1

2

2

1

2

4

4

2

3

8

8

3

4

16

16

4

  • Notice that the tables are mirrors of each other

    • The inputs and outputs are swapped

    • This is exactly the inverse relationship at work

How do ordered pairs relate between the two functions?

  • If (s,t) is a point on the graph of the exponential function g(x)=bx

    • then (t,s) is a point on the graph of the logarithmic function  f(x)=logbx

  • This follows directly from the inverse relationship

    • if bs=t, then logbt=s

  • E.g. since 23=8

    • the point (3,8) is on the graph of g(x)=2x

    • and the point (8,3) is on the graph of  f(x)=log2x

Graphs of logarithmic functions

  • The graph of  f(x)=logbx is the reflection of the graph of g(x)=bx over the line  y=x

    • This is true for any base b (with b>0 and b1)

    • It follows from the general principle that the graph of any function and its inverse are reflections of each other over  y=x

  • Note how features of the graphs swap under reflection:

    • The exponential function has a horizontal asymptote ( y=0)

      • the logarithmic function has a vertical asymptote (x=0)

    • The exponential function passes through (0,1)

      • the logarithmic function passes through (1,0)

    • The exponential function has domain all real numbers and range y>0

      • the logarithmic function has domain x>0 and range all real numbers

    • These graph features are true for any base b (with b>0 and b1)

Graph showing lines y=x, y=2^x, and y=log_2(x). Asymptotes at y=0 and x=0, points at (0,1) and (1,0).
Graphs of exponential and logarithmic functions

Worked Example

The exponential function g(x)=3x and the logarithmic function  f(x)=log3x are inverse functions.

(a) Without using a calculator, complete the tables of values for g(x)=3xand  f(x)=log3x.

(b) Show that  f(g(x))=x by evaluating  f(g(2)).

x

2

1

0

1

2

3

g(x)=3x

x

19

13

1

3

9

27

f(x)=log3x

Answer:

(a)

You can use rules of exponents to complete the table for g

32=132=19             31=131=13             30=1

31=3             32=9             33=27

x

2

1

0

1

2

3

g(x)=3x

19

13

1

3

9

27

To find the table for  f(x)=log3x, swap the inputs and outputs in the table for g

  • This works because the functions are inverses

x

19

13

1

3

9

27

 f(x)=log3x

2

1

0

1

2

3

(b)

Start with the value of g(2)

g(2)=32=9

Substitute that into  f(x), and find the value

 f(g(2))=f(9)=log39=2

So the output is equal to the input

The value of  f(g(2)) equals the original input (2), illustrating the fact that  f(g(x))=x

Rewriting exponential functions using logarithms

How can logarithms be used to rewrite exponential expressions?

  • Logarithms provide a way to change the base of an exponential expression

    • This can be useful for comparing or simplifying expressions

  • The key identity is bx=c(logcb)(x)

    • This works because b=clogcb

      • I.e., by the inverse properties of logarithms and exponentials

      • c 'cancels' logc

    • Therefore bx=(clogcb)x=c(logcb)(x)

Examiner Tips and Tricks

The identity bx=c(logcb)(x) may look complicated

  • But it follows directly from the fact that b can be written as clogcb

  • If you can remember that one fact, you can derive the rest

Also be careful with the notation here!

  • (logcb)(x) means

    • the number logcb

    • times the input variable x

  • It does not mean logc(bx)

    • which is the logarithm to base c of bx

  • This means any exponential function can be rewritten using any base you choose

    • With the notation used above, you can use the identity to rewrite an exponential with base b as an exponential with base c

  • E.g. rewrite 3x using the base e

    • Use the identity

      • 3x=e(ln3)(x)

        • since ln3=loge3

    • Or with the brackets expanded

      • 3x=exln3

  • E.g. rewrite 5x using base 10

    • Using the identity

      • 5x=10(log105)(x)

    • Or with the brackets expanded

      • 5x=10xlog5

Examiner Tips and Tricks

A very common application is converting any exponential to base e.

  • For this the formula can be written more simply as bx=exlnb

When rewriting exponential expressions in a free response question on the exam, always show the intermediate steps clearly.

Worked Example

Rewrite each of the following exponential expressions in the specified form.

(a) Rewrite 7x in the form ekx, where k is a constant. Express k as a decimal approximation to three decimal places.

(b) Rewrite 4x in the form 2mx, where m is a constant.

Answer:

(a)

Use the identity bx=e(lnb)(x) with b=7

7x=e(ln7)(x)

Use a calculator to find the value of ln7

k=ln7=1.945910...

Round k to 3 decimal places

7x=e1.946x

(b)

The easiest way to do this is to substitute22 in place of 4

  • and then use rules of exponents

4x=(22)x=22x

That is in the form you are looking for, with m=2

You could also work this out using the identity bx=c(logcb)(x)

  • with b=4 and c=2

4x=2(log24)(x)=2(2)(x)=22x

  • That follows since log24=2

4x=22x

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.