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College Algebra : Logarithmic Functions

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Example Questions

Example Question #1 : Logarithmic Functions

Solve the following for x:

$\log(2x+4)=1$

Possible Answers:

$\frac{-3}{2}$

Correct answer:

Explanation:

To solve, you must first "undo" the log. Since no base is specified, you assume it is 10. Thus, we need to take 10 to both sides.

$10^{\log(2x+4)}=10^1$

2x+4=10

Now, simply solve for x.

2x=6

x=3

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Example Question #2 : Logarithmic Functions

Simplify the following:

$\ln{2x}+\ln{y^2}-\ln{z^7}$

Possible Answers:

$\ln{\frac{z^7}{2xy^2}}$

$\ln{\frac{2xy^2}{z^7}}$

$\ln{\frac{y^2z^7}{2x}}$

$\ln{\frac{2xz^7}{y^2}}$

Correct answer:

$\ln{\frac{2xy^2}{z^7}}$

Explanation:

To solve, you must combine the logs into 1 log, instead of three separate ones. To do this, you must remember that when adding logs, you multiply their insides, and when you subtract them, you add their insides. Therefore,

$\ln{2x}+\ln{y^2}-\ln{z^7}$

$\ln{2xy^2}-\ln{z^7}$

$\ln{\frac{2xy^2}{z^7}}$

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Example Question #12 : Exponential And Logarithmic Functions

Solve for y in the following expression:

$\log(y) + \log(5x) = 2$

Possible Answers:

y = 20x

$y = 2 - \log(5x)$

y=100-5x

$y = \frac{20}{x}$

Correct answer:

$y = \frac{20}{x}$

Explanation:

To solve for y we first need to get rid of the logs.

$\log(y)+\log(5x) = 2 \rightarrow 10^{\log(y)} \cdot 10^{\log(5x)} = 100$

Then we get $y \cdot 5x = 100$ .

After that, we simply have to divide by 5x on both sides:

$y = \frac{100}{5x} = \frac{20}{x}$

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Example Question #13 : Exponential And Logarithmic Functions

Solve for .

ln(x^2 -3)=0

Possible Answers:

$x=\pm{2}$

$x=\pm{4}$

$x=\pm{\sqrt{2}}$

x={2}

x=-2

Correct answer:

$x=\pm{2}$

Explanation:

To solve this natural logarithm equation, we must eliminate the ln() operation. To do that, we must remember that ln() is simply log with base . So, we raise both side of the equation to the power.

$e^{ln(x^2 -3)}=e^0$

This simplifies to

x^2-3=1 . Remember that anything raised to the 0 power is 1.

Continuing to solve for x,

x^2=4

$x=\pm2$

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Example Question #14 : Exponential And Logarithmic Functions

Solve for .

$log_{7}(7x+7)=1$

Possible Answers:

x=-7

x=0

x=7

x=1

$x=\sqrt{1}$

Correct answer:

x=0

Explanation:

To eliminate the log() operation, simply raise both side of the equation to the 7th power because the base of the log() operation is 7.

$7^{log_{7}(7x+7)}=7^{1}$

This simplifies to

7x+7=7

7x=0

x=0

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Example Question #12 : College Algebra

N > 0 ; $a \in (0, 1) \cup (1, \infty)$

True or false:

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

if and only if either b= 10 or b= e .

Possible Answers:

False

True

Correct answer:

False

Explanation:

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

is a direct statement of the Change of Base Property of Logarithms. If N > 0 and $a \in (0, 1) \cup (1, \infty)$ , this property holds true for any $b \in (0, 1) \cup (1, \infty)$ - not just b = e, 10 .

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Example Question #13 : College Algebra

Evaluate $\log_{-1} (-1)$

Possible Answers:

$\log_{-1} (-1) = 1$ is an undefined quantity.

Correct answer:

$\log_{-1} (-1) = 1$ is an undefined quantity.

Explanation:

$\log_{-1} (-1)$ is undefined for two reasons: first, the base of a logarithm cannot be negative, and second, a negative number cannot have a logarithm.

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Example Question #7 : Logarithmic Functions

Use the properties of logarithms to rewrite as a single logarithmic expression:

$- 2+ t \log 2$

Possible Answers:

$\log \frac{t^{2} }{100}$

$\log( 2^{t} -100)$

$\log(2t -100)$

$\log \frac{2^{t} }{100}$

$\log \frac{t}{50}$

Correct answer:

$\log \frac{2^{t} }{100}$

Explanation:

$b \log M = \log M^{b}$ , so

$- 2+ t \log 2$

$= - 2+ \log 2^{t}$

$N= \log 10^{N}$ , so the above becomes

$= \log 10^{-2}+ \log 2^{t}$

$= \log \frac{1}{100}+ \log 2^{t}$

$\log M + \log N = \log MN$ , so the above becomes

$=\log\left ( \frac{1}{100} \cdot 2^{t} \right )$

$=\log \frac{2^{t} }{100}$

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Example Question #8 : Logarithmic Functions

Use the properties of logarithms to rewrite as a single logarithmic expression:

$\frac{2+ \log_{5}t}{\log_{5}17}$

Possible Answers:

$\log_{17}(25 +t)$

$\log_{17}(32 +t)$

$\log_{17}32 t$

$\log_{17}25 t$

None of the other choices gives the correct response.

Correct answer:

$\log_{17}25 t$

Explanation:

$N= \log_{a} a^{N}$ , so

$\frac{2+ \log_{5}t}{\log_{5}17}$

$= \frac{ \log_{5}5^{2}+ \log_{5}t}{\log_{5}17}$

$= \frac{ \log_{5}25+ \log_{5}t}{\log_{5}17}$

$\log_{a} M + \log_{a} N = \log_{a} MN$ , so the above becomes

$= \frac{ \log_{5}25 t}{\log_{5}17}$

By the Change of Base Property,

$\frac{\log_{a}M}{\log_{a}N} = \log_{N}M$ , so the above becomes

$= \log_{17}25 t$ ,

the correct response.

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Example Question #2 : Logarithmic Functions

Expand the logarithm: $\ln \sqrt[3]{\frac{x}{y}}$

Possible Answers:

$\frac{1}{3} ( \ln x +\ln y)$

$\ln x +3\ln y$

$3 \ln x -\ln y$

None of these

$\frac{1}{3} ( \ln x -\ln y)$

Correct answer:

$\frac{1}{3} ( \ln x -\ln y)$

Explanation:

We expand this logarithm based on the property: $\ln \frac{u}{v}= \ln u-\ln v$

and $\ln u^{n}= n \ln u$ .

$\ln \sqrt[3]{\frac{x}{y}}\rightarrow\frac{1}{3}\ln\frac{x}{y}\rightarrow \boldsymbol{\frac{1}{3}(\ln x-\ln y)}$

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College Algebra : Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Logarithmic Functions

Example Question #2 : Logarithmic Functions

Example Question #12 : Exponential And Logarithmic Functions

Example Question #13 : Exponential And Logarithmic Functions

Example Question #14 : Exponential And Logarithmic Functions

Example Question #12 : College Algebra

Example Question #13 : College Algebra

Example Question #7 : Logarithmic Functions

Example Question #8 : Logarithmic Functions

Example Question #2 : Logarithmic Functions

All College Algebra Resources

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