College Algebra - Logarithmic Functions

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

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

$\log_{17}25 t$

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

$\log_{17}32 t$

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

None of the other choices gives the correct response.

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.

Simplify the following:

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

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

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

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

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

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,

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

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

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

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

$4\ln x+\frac{1}{2}\ln y-6\ln z$

$4\ln x+\frac{1}{2}\ln y+6\ln z$

$4\ln x-\frac{1}{2}\ln y-6\ln z$

$4\ln x+2\ln y-6\ln z$

$4\ln x+2\ln y+6\ln z$

Explanation

We expand this logarithm based on the following properties:

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

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

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

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

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

$- 2+ t \log 2$

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

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

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

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

$\log(2t -100)$

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

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

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

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

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

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

$\ln x +3\ln y$

None of these

$3 \ln x -\ln y$

Explanation

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

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

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

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

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

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

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

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

None of these

Explanation

We condense this logarithm based on the following properties:

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

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

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

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

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

Solve for .

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

$x=\sqrt{1}$

Explanation

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

Solve for y in the following expression:

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

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

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

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:

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

Solve for .

$x=\pm{2}$

$x=\pm{4}$

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

Explanation

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,

$x=\pm2$

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

True or false:

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

if and only if either or .

False

True

Explanation

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

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 .

Logarithmic Functions

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