Logarithmic Functions - College Algebra

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Question

Solve the following for x:

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

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Answer

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.

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

Now, simply solve for x.

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Question

Simplify the following:

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

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Answer

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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Question

Solve for y in the following expression:

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

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Answer

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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Question

Solve for .

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Answer

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

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

This simplifies to

. Remember that anything raised to the 0 power is 1.

Continuing to solve for x,

$x=\pm2$

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Question

Solve for .

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

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Answer

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

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

This simplifies to

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Question

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

True or false:

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

if and only if either or .

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Answer

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

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

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Question

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

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Answer

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

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

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

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Answer

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

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

$- 2+ t \log 2$

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Answer

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

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

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Answer

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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Question

Expand this logarithm: $\ln \frac{x^4\sqrt{y}}{z^6}$

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Answer

We expand this logarithm based on the following properties:

$\ln \frac{u}{v}= \ln u-\ln v$

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

$\ln uv=\ln u+\ln v$

$\ln \frac{x^4\sqrt{y}}{z^6}\rightarrow \ln x^4\ln\sqrt{y}-\ln z^6\rightarrow \boldsymbol{4\ln x+\frac{1}{2}\ln y-6\ln z}$

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Question

Condense this logarithm: $\frac{1}{3}[2\ln (x+3)+\ln x-\ln(x^2-1)]$

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Answer

We condense this logarithm based on the following properties:

$\ln \frac{u}{v}= \ln u-\ln v$

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

$\ln uv=\ln u+\ln v$

$\frac{1}{3}[2\ln (x+3)+\ln x-\ln(x^2-1)]\rightarrow \frac{1}{3}[\ln(x+3)^2+\ln x-\ln(x^2-1)]\rightarrow \frac{1}{3}[ \ln x(x+3)^2-\ln (x^2-1)]\rightarrow$

$\frac{1}{3}[\ln\frac{x(x+3)^2}{x^2-1}]\rightarrow\boldsymbol{ \ln \sqrt[3]{\frac{x(x+3)^2}{x^2-1}}}$

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Question

Solve for y in the following expression:

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

Tap to reveal answer

Answer

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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Question

Solve the following for x:

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

Tap to reveal answer

Answer

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.

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

Now, simply solve for x.

← Didn't Know|Knew It →

Question

Simplify the following:

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

Tap to reveal answer

Answer

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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Question

Solve for .

Tap to reveal answer

Answer

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

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

This simplifies to

. Remember that anything raised to the 0 power is 1.

Continuing to solve for x,

$x=\pm2$

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Question

Solve for .

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

Tap to reveal answer

Answer

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

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

This simplifies to

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Question

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

True or false:

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

if and only if either or .

Tap to reveal answer

Answer

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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