Free preview

4.3 Logarithmic Functions

Connect logarithms to exponents.

Before you solve

Ideas in this lesson

Explore the lesson brief and the worked example for Idea $1$ . Create a free account to continue with the remaining ideas and practice.

Lesson brief and rules

4.3 Logarithmic Functions

$\log _b(x) = y$ means $b^{y} = x$ , where $b > 0, b \ne 1$ , and $x > 0$ .

Remember
A real logarithm requires a positive argument.

Worked example

1

01 / Complete the exact common-

\log

table for arguments

\dfrac{1}{1000}

\dfrac{1}{100}

\dfrac{1}{10}

1

\sqrt{10}

10

100

1000

. Convert

2^{u}=v

and

\log_{2}(3z)=w

into the other form, stating argument restrictions. Convert each of these six given statements into its other form and reverse the conversion to check it:

Row	Given statement	Required conversion
$1$	$\log_{2}(4)=2$	Write the equivalent form
$2$	$3^{-1}=\dfrac{1}{3}$	Write the equivalent form
$3$	$\log_{10}(1)=0$	Write the equivalent form
$4$	$4^{\dfrac{3}{2}}=8$	Write the equivalent form
$5$	$\log_{2}(\sqrt{2})=\dfrac{1}{2}$	Write the equivalent form
$6$	$4^{- \dfrac{3}{2}}=\dfrac{1}{8}$	Write the equivalent form

Also evaluate

e^{2\ln(3)}

exactly using an inverse identity. Evaluate

\log_{2}(\dfrac{1}{8})

exactly and convert it to exponential form. State and explain both inverse identities and their domains. Find the valid base in

\log_a(4)=2

and rewrite

\ln u=-3

exponentially. Define common/natural logarithms and evaluate

\dfrac{\ln{\left(\dfrac{3}{2} \right)}}{\ln{\left(10 \right)}}

\ln{\left(\sqrt{2} \right)}

\ln{\left(1 + \sqrt{3} \right)}

to four decimals. Show the requested work; written explanations and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

02 /

\log_a u=w\iff a^{w}=u

requires

a>0,\ a\ne1,\ u>0

. The base stays fixed; the logarithm gives the exponent. The exponential is one-to-one, so its inverse exists.

03 /

\log_{2}(\dfrac{1}{8})=-3

because

2^{-3}=\dfrac{1}{8}

04 /

\log_{2}(2^{x})=x

holds for every real input, whereas

2^{\log_{2}x}=x

requires

x>0

05 /

\log_a(4)=2\iff a^{2}=4

. Raising the positive argument to the reciprocal power gives

a=(4)^{\dfrac{1}{2}}=2

. Retain only bases

a>0,\ a\ne1

06 /

\ln u=-3\iff u=e^{-3}

. The natural logarithm has base

e

; the common logarithm has base

10

07 / Numerical evaluations, with the full argument grouped before rounding:

\dfrac{\ln{\left(\dfrac{3}{2} \right)}}{\ln{\left(10 \right)}}\approx0.1761

\ln{\left(\sqrt{2} \right)}\approx0.3466

\ln{\left(1 + \sqrt{3} \right)}\approx1.0051

08 /

2^{u}=v

is equivalent to

\log_{2}v=u

with

v>0

. Also

\log_{2}(3z)=w

is equivalent to

3z=2^{w}

, requiring

z>0

. These symbolic conversions preserve the whole argument, not just its variable.

09 / The exact common-

\log

table follows from

\log_{10}(10^{j})=j

Argument	Common logarithm
$\dfrac{1}{1000}$	$-3$
$\dfrac{1}{100}$	$-2$
$\dfrac{1}{10}$	$-1$
$1$	$0$
$\sqrt{10}$	$\dfrac{1}{2}$
$10$	$1$
$100$	$2$
$1000$	$3$

10 /

Row	Logarithmic form	Equivalent exponential form
$1$	$\log_{2}(4)=2$	$2^{2}=4$
$2$	$\log_{3}(\dfrac{1}{3})=-1$	$3^{-1}=\dfrac{1}{3}$
$3$	$\log_{10}(1)=0$	$10^{0}=1$
$4$	$\log_{4}(8)=\dfrac{3}{2}$	$4^{\dfrac{3}{2}}=8$
$5$	$\log_{2}(\sqrt{2})=\dfrac{1}{2}$	$2^{\dfrac{1}{2}}=\sqrt{2}$
$6$	$\log_{4}(\dfrac{1}{8})=- \dfrac{3}{2}$	$4^{- \dfrac{3}{2}}=\dfrac{1}{8}$

11 /

e^{2\ln(3)}=(e^{\ln(3)})^{2}=3^{2}=9

. The positive argument permits the inverse identity before applying the integer power.

Worked example 2 · Use the inverse definition and evaluate logarithms

Complete the exact common- $\log$ table for arguments $\dfrac{1}{1000}$ , $\dfrac{1}{100}$ , $\dfrac{1}{10}$ , $1$ , $\sqrt{10}$ , $10$ , $100$ , $1000$ . Convert $3^{u}=v$ and $\log_{3}(3z)=w$ into the other form, stating argument restrictions. Convert each of these six given statements into its other form and reverse the conversion to check it:

Row	Given statement	Required conversion
$1$	$\log_{3}(9)=2$	Write the equivalent form
$2$	$4^{-1}=\dfrac{1}{4}$	Write the equivalent form
$3$	$\log_{10}(1)=0$	Write the equivalent form
$4$	$4^{\dfrac{3}{2}}=8$	Write the equivalent form
$5$	$\log_{3}(\sqrt{3})=\dfrac{1}{2}$	Write the equivalent form
$6$	$4^{- \dfrac{3}{2}}=\dfrac{1}{8}$	Write the equivalent form

Also evaluate

e^{3\ln(4)}

exactly using an inverse identity. Evaluate

\log_{3}(\dfrac{1}{9})

exactly and convert it to exponential form. State and explain both inverse identities and their domains. Find the valid base in

\log_a(27)=3

and rewrite

\ln u=-1

exponentially. Define common/natural logarithms and evaluate

\dfrac{\ln{\left(2 \right)}}{\ln{\left(10 \right)}}

\ln{\left(\sqrt{3} \right)}

\ln{\left(3 \right)}

$\log_a u=w\iff a^{w}=u$ requires $a>0,\ a\ne1,\ u>0$ . The base stays fixed; the logarithm gives the exponent. The exponential is one-to-one, so its inverse exists.
$\log_{3}(\dfrac{1}{9})=-2$ because $3^{-2}=\dfrac{1}{9}$ .
$\log_{3}(3^{x})=x$ holds for every real input, whereas $3^{\log_{3}x}=x$ requires $x>0$ .
$\log_a(27)=3\iff a^{3}=27$ . Raising the positive argument to the reciprocal power gives $a=(27)^{\dfrac{1}{3}}=3$ . Retain only bases $a>0,\ a\ne1$ .
$\ln u=-1\iff u=e^{-1}$ . The natural logarithm has base $e$ ; the common logarithm has base $10$ .
Numerical evaluations, with the full argument grouped before rounding: $\dfrac{\ln{\left(2 \right)}}{\ln{\left(10 \right)}}\approx0.3010$ , $\ln{\left(\sqrt{3} \right)}\approx0.5493$ , $\ln{\left(3 \right)}\approx1.0986$ .
$3^{u}=v$ is equivalent to $\log_{3}v=u$ with $v>0$ . Also $\log_{3}(3z)=w$ is equivalent to $3z=3^{w}$ , requiring $z>0$ . These symbolic conversions preserve the whole argument, not just its variable.
The exact common- $\log$ table follows from $\log_{10}(10^{j})=j$ .

Argument	Common logarithm
$\dfrac{1}{1000}$	$-3$
$\dfrac{1}{100}$	$-2$
$\dfrac{1}{10}$	$-1$
$1$	$0$
$\sqrt{10}$	$\dfrac{1}{2}$
$10$	$1$
$100$	$2$
$1000$	$3$

Row	Logarithmic form	Equivalent exponential form
$1$	$\log_{3}(9)=2$	$3^{2}=9$
$2$	$\log_{4}(\dfrac{1}{4})=-1$	$4^{-1}=\dfrac{1}{4}$
$3$	$\log_{10}(1)=0$	$10^{0}=1$
$4$	$\log_{4}(8)=\dfrac{3}{2}$	$4^{\dfrac{3}{2}}=8$
$5$	$\log_{3}(\sqrt{3})=\dfrac{1}{2}$	$3^{\dfrac{1}{2}}=\sqrt{3}$
$6$	$\log_{4}(\dfrac{1}{8})=- \dfrac{3}{2}$	$4^{- \dfrac{3}{2}}=\dfrac{1}{8}$

e^{3\ln(4)}=(e^{\ln(4)})^{3}=4^{3}=64

. The positive argument permits the inverse identity before applying the integer power.
Written-work review:

Complete all eight exact common- $\log$ values and all six bidirectional log/exponential conversion rows.
Preserve the base, exponent and positive argument and state the unequal domains of the two inverse identities.
Solve the unknown base/argument while rejecting invalid bases and evaluate the scaled- $\log$ inverse expression exactly.
Evaluate the separate common/natural logarithms with grouped arguments and four-decimal rounding.

Read the graphs

Build logarithm graphs and recover their bases

Include a table and curve for twice the logarithm plus one. For base

2

, choose five powers of the base to make a logarithm table. Draw the exponential, its logarithmic inverse and their reflection line together. Explain the exchanged domain/range and coordinates. Recover the base of a separate logarithm graph through

(\sqrt{3},\dfrac{1}{2})

. Reference plot. Curve

1

: unknown logarithm.

Show the requested work; written explanations and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

Input	Logarithm	Twice the logarithm plus one
$\dfrac{1}{4}$	$-2$	$-3$
$\dfrac{1}{2}$	$-1$	$-1$
$1$	$0$	$1$
$2$	$1$	$3$
$4$	$2$	$5$

a^{\dfrac{1}{2}}=\sqrt{3}

at the marked point gives the positive base

a=3

.
The logarithm has domain

(0,\infty)

, range

\mathbb{R}

, vertical asymptote

x=0

and intercept

(1,0)

. The exponential has domain all reals, positive range and intercept

(0,1)

.
Reflection in

y=x

exchanges coordinates: each

(r,s)

on the exponential gives

(s,r)

on the logarithm. A base between 0 and 1 makes both curves decreasing; a base above 1 makes both increasing.
Reference plot. Curve 1:

2^{x}

. Curve 2:

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

. Curve 3:

x

Reference plot. Curve 1:

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

. Curve 2:

\dfrac{2 \ln{\left(x \right)}}{\ln{\left(2 \right)}} + 1

The additional transformed logarithm doubles every base-log output and then adds 1. Its domain remains positive and its range remains all real numbers.
Written-work review:

Choose powers of the base to complete both logarithm tables and draw the transformed curve.
Recover the unknown logarithm base from the actual marked graph and its point.
Draw the exponential, logarithmic inverse and reflection line, exchanging coordinates and domains/ranges correctly.

Answer: $a=3$ ; inverse reflection exchanges coordinates.

Transform logarithms and distinguish absolute values

Match these four graphs to

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

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

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

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

, explaining each input/output reflection. Graph

A

. Reference plot. Curve

1

: unidentified curve.

Graph

B

. Reference plot. Curve

1

: unidentified curve.

Graph

C

. Reference plot. Curve

1

: unidentified curve.

Graph

D

. Reference plot. Curve

1

: unidentified curve.

Use transformations to sketch

f(x)=\dfrac{\ln{\left(x + 3 \right)}}{\ln{\left(2 \right)}} + 1

. Give domain, range, vertical asymptote and both tail directions, plus both intercepts or a reason one does not exist. Show the requested work; written explanations and sketches require teacher review. Also test whether input

\dfrac{2}{3}

is admissible, evaluating the output to three decimals if it is. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

Start with the base logarithm; translate horizontally by $-3$ , then translate vertically by $1$ .
$x + 3>0$ gives domain $(-3,\infty)$ . Range is all real numbers and the vertical asymptote is $x=-3$ .
At the domain boundary, the argument approaches zero from above. The output tends to negative infinity. At the infinite end of the domain it tends to the opposite infinity.
For the horizontal intercept, isolate the logarithm at $-1.0$ , exponentiate and solve: $x=-2.5000$ . The vertical intercept is approximately $(0,2.5850)$ .
Reference plot. Curve $1$ : $\dfrac{\ln{\left(x + 3 \right)}}{\ln{\left(2 \right)}} + 1$ .

Reflection matching: Graph A:

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

; Graph B:

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

; Graph C:

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

; Graph D:

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

. A minus sign in the input reflects across the vertical axis; a minus sign outside reflects across the horizontal axis; using both produces both reflections.
The extra input is admissible and gives

2.874

.
Written-work review:

Identify all four basic reflection graphs, including simultaneous input and output reflections.
Apply the shifts/reflections in the given transformed formula and give the strict domain and real range.
State the vertical asymptote, both boundary/tail directions and both intercepts or their nonexistence.
Check the extra test input against the original domain and evaluate to three decimals only if admissible.

Answer: Domain $(-3,\infty)$ ; vertical asymptote $x=-3$ . The extra input is admissible and gives $2.874$ .

Prove large-scale facts and derive logarithmic secants

Use the marked graph of

y=\ln x

and its secant below to infer the coordinates of

A

and

B

. The horizontal coordinate of

B

is marked

e^{a}

;

A

is the horizontal intercept. The reference diagram uses

a=1

. Derive the symbolic secant equation for any real

a\ne0

, explain the excluded parameter, and reproduce the reference sketch. Reference plot. Curve

1

y = \ln x

. Curve

2

: secant through

A

and

B

The logarithm graph crosses the horizontal axis at $(1,0)$ . Its point with horizontal coordinate $e^{a}$ has vertical coordinate $a$ because $\ln(e^{a})=a$ .
$m=\dfrac{a-0}{e^{a}-1}$ for $a\ne0$ . Point-slope form gives $y=\dfrac{a}{e^{a}-1}(x-1)$ .
This is a secant through two distinct points, not a tangent. At $a=0$ the displayed quotient is undefined and the two points coincide.
For $a=1$ , slope is approximately $0.5820$ .
Reference plot. Curve $1$ : $\ln{\left(x \right)}$ . Curve $2$ : $\dfrac{x - 1}{-1 + e}$ .

Written-work review:

Show the requested logarithmic identity or inequality argument rather than relying only on decimal evidence.
Infer the two distinct points from the marked logarithm graph, derive the secant, retain the nonzero parameter restriction and do not substitute a tangent formula.

Answer: $y=\dfrac{a}{e^{a}-1}(x-1)$ for $a\ne0$ .

15 questions with a worked answer for each.Start practice ↗

Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

Read this lesson preview

MathLab with Mr. Ahmed

4.3 Logarithmic Functions

Ideas in this lesson

4.3 Logarithmic Functions

Read the graphs

Build logarithm graphs and recover their bases

Transform logarithms and distinguish absolute values

Prove large-scale facts and derive logarithmic secants