Change the base of a logarithm Algebra I

Change the base of the following logarithms.

\log_4\left(62\right)

\dfrac{\log_2\left(62\right)}2

\dfrac{\log_2\left(62\right)}{4}

\log_2\left(62\right)-2

\log_2\left(62\right)-4

\log_9\left(73\right)

\dfrac{\log_3\left(73\right)}{9}

\dfrac{\log_3\left(73\right)}2

\log_3\left(73\right)-2

\log_3\left(73\right)-9

\log_2\left(\sqrt{e}\right)

\dfrac{2}{\ln\left(2\right)}

\dfrac{1}{\sqrt{\ln\left(2\right)}}

\dfrac{1}{2}-\ln\left(2\right)

\dfrac{1}{2\ln\left(2\right)}

\log_\dfrac{1}{100}\left(253\right)

-\dfrac{1}{10}\log\left(253\right)

\log\left(253\right)-2

-\dfrac{1}{2}\log\left(253\right)

-2\log\left(253\right)

\log_\sqrt[3]{5}\left(127\right)

3\log_5\left(127\right)

\dfrac{\log_5\left(127\right)}{3}

\log_5\left(127\right)-3

-\dfrac{\log_5\left(127\right)}{3}

\log_8 \left(50\right)

\log_2\left(50\right)-3

\log_2\left(50\right)-27

\dfrac{\log_2\left(50\right)}{3}

\dfrac{\log_2\left(50\right)}{27}

\log_\dfrac{1}{\sqrt{e}} \left(12\right)

-\dfrac{1}{2}\ln\left(12\right)

-2\ln\left(12\right)

2\ln\left(12\right)

\dfrac{1}{2}\ln\left(12\right)