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College Algebra : Logarithmic Functions

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Example Question #11 : Logarithmic Functions

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We expand this logarithm based on the following properties:

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

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

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

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

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Example Question #21 : College Algebra

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

Possible Answers:

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

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

None of these

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

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

Correct answer:

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

Explanation:

We condense this logarithm based on the following properties:

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

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

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

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

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

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College Algebra : Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

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Example Question #11 : Logarithmic Functions

Example Question #21 : College Algebra

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