College Algebra : Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #1 : Logarithmic Functions

Solve the following for x:

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

Possible Answers:

\displaystyle 3

\displaystyle \frac{-3}{2}

\displaystyle 7

\displaystyle 48

Correct answer:

\displaystyle 3

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.

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

\displaystyle 2x+4=10

Now, simply solve for x.

\displaystyle 2x=6

\displaystyle x=3

Example Question #2 : Logarithmic Functions

Simplify the following:

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

Possible Answers:

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

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

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

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

Correct answer:

\displaystyle \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,

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

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

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

Example Question #3 : Logarithmic Functions

Solve for y in the following expression:

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

Possible Answers:

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

\displaystyle y=100-5x

\displaystyle y = 20x

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

Correct answer:

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

Explanation:

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

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

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

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

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

Example Question #3 : Logarithmic Functions

Solve for \displaystyle x.

\displaystyle ln(x^2 -3)=0

Possible Answers:

\displaystyle x=\pm{4}

\displaystyle x={2}

\displaystyle x=-2

\displaystyle x=\pm{2}

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

Correct answer:

\displaystyle x=\pm{2}

Explanation:

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

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

This simplifies to

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

Continuing to solve for x,

\displaystyle x^2=4

\displaystyle x=\pm2

Example Question #14 : College Algebra

Solve for \displaystyle x.

\displaystyle log_{7}(7x+7)=1

Possible Answers:

\displaystyle x=0

\displaystyle x=-7

\displaystyle x=\sqrt{1}

\displaystyle x=7

\displaystyle x=1

Correct answer:

\displaystyle x=0

Explanation:

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

\displaystyle 7^{log_{7}(7x+7)}=7^{1}

This simplifies to 

\displaystyle 7x+7=7

\displaystyle 7x=0

\displaystyle x=0

Example Question #4 : Logarithmic Functions

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

True or false:

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

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

Possible Answers:

True

False

Correct answer:

False

Explanation:

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

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

Example Question #5 : Logarithmic Functions

Evaluate \displaystyle \log_{-1} (-1)

Possible Answers:

\displaystyle 1

\displaystyle e

\displaystyle 0

\displaystyle -1

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

Correct answer:

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

Explanation:

\displaystyle \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. 

Example Question #7 : Logarithmic Functions

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

\displaystyle - 2+ t \log 2

Possible Answers:

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

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

\displaystyle \log(2t -100)

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

\displaystyle \log \frac{t}{50}

Correct answer:

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

Explanation:

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

\displaystyle - 2+ t \log 2

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

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

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

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

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

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

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

Example Question #8 : Logarithmic Functions

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

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

Possible Answers:

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

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

\displaystyle \log_{17}32 t

\displaystyle \log_{17}25 t

None of the other choices gives the correct response.

Correct answer:

\displaystyle \log_{17}25 t

Explanation:

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

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

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

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

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

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

By the Change of Base Property,

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

\displaystyle = \log_{17}25 t,

the correct response.

Example Question #2 : Logarithmic Functions

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

Possible Answers:

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

\displaystyle \ln x +3\ln y

\displaystyle 3 \ln x -\ln y

None of these

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

Correct answer:

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

Explanation:

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

and \displaystyle \ln u^{n}= n \ln u.

 

\displaystyle \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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