Properties of Logs

1. Evaluate each of the following expressions using the propeties of logs (and no calculator)

$log_{3}\sqrt[3]{81}=log_{3}\left&space;(&space;81&space;\right&space;)^{\frac{1}{3}}=log_{3}\left&space;(&space;3^{4}&space;\right&space;)^{\frac{1}{3}}=log_{3}3^{\frac{4}{3}}=\frac{4}{3}\cdot&space;log_{3}3=\frac{4}{3}$

$ln\left&space;(&space;lne^{e^{200}}&space;\right&space;)=ln\left&space;(&space;e^{200}\cdot&space;lne&space;\right&space;)=lne^{200}=200\cdot&space;lne=200$

2. Use the properties of logs to expand the following expressions.

$log_{5}\sqrt[4]{x^{3}\left&space;(&space;x^{2}+1&space;\right&space;)}=log_{5}\left&space;(&space;\sqrt[4]{x^{3}}&space;\right&space;)\left&space;(&space;\sqrt[4]{x^{2}+1}&space;\right&space;)=log_{5}x^{\frac{3}{4}}\cdot&space;\left&space;(&space;x^{2}+1&space;\right&space;)^{\frac{1}{4}}=log_{5}x^{\frac{3}{4}}+log_{5}\left&space;(&space;x^{2}+1&space;\right&space;)^{\frac{1}{4}}=\frac{3}{4}log_{5}x+\frac{1}{4}log_{5}(x^{2}+1)$

$log\sqrt{x\sqrt{y\sqrt{z}}}=log\sqrt{x\sqrt{y\cdot&space;z^{\frac{1}{2}}}}=log\sqrt{x\cdot&space;\left&space;(&space;yz^{}\frac{1}{2}&space;\right&space;)^{\frac{1}{2}}}=log\sqrt{xy^{\frac{1}{2}}z^{\frac{1}{4}}}=logx^{\frac{1}{2}}y^{\frac{1}{4}}z^{\frac{1}{8}}=\frac{1}{2}logx+\frac{1}{4}logy+\frac{1}{8}logz$

3. Use the properties of logs to condense the following expressions

$4lnx-\frac{1}{3}ln\left&space;(&space;x^{2}+1&space;\right&space;)+2ln(x-1)=lnx^{4}-ln\left&space;(&space;x^{2}+1&space;\right&space;)^{\frac{1}{3}}+ln(x-1)^{2}=lnx^{4}+ln(x-1)^{2}-ln\sqrt[3]{x^{2}+1}=lnx^{4}(x-1)^{2}-ln\sqrt[3]{x^{2}+1}=ln\frac{x^{4}(x-1)^{2}}{\sqrt[3]{x^{2}+1}}$

$log(x^{2}-1)-ln(x-1)=\frac{ln(x^{2}-1)}{ln10}-ln(x-1)=\frac{1}{ln10}\cdot&space;ln(x^{2}-1)-ln(x-1)=ln(x^{2}-1)^{\frac{1}{ln10}}-ln(x-1)=ln\frac{(x^{2}-1)^{\frac{1}{ln10}}}{x-1}$

4. If  $log_{7}x=Alog_{\frac{2}{3}}x$, use the change of base formula to find the value of A.

$log_{7}x=Alog_{\frac{2}{3}}x\Rightarrow&space;A=\frac{log_{7}x}{log_{\frac{2}{3}}x}=\frac{\frac{logx}{log7}}{\frac{logx}{log\frac{2}{3}}}=\frac{log\frac{2}{3}}{log7}=log_{7}\frac{2}{3}$

5. Simplify the following to a single log expression of the form logba:

$\left&space;(log_{7}3&space;\right&space;)\left&space;(&space;log_{2}5&space;\right&space;)\left&space;(&space;log_{5}7&space;\right&space;)=\frac{log3}{log7}\cdot&space;\frac{log5}{log2}\cdot&space;\frac{log7}{log5}=\frac{log3}{log2}=log_{2}3$