## Logarithms

${\mathrm{log}}_{2}\left(32\right)=x$

This equation simply asks the question 2 raised to what power is 32. It could be rewritten as:

${2}^{x}=\mathrm{32}$

## Logarithmic Identities

• ${\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)={\mathrm{log}}_{b}\left(xy\right)$
• ${\mathrm{log}}_{b}\left(x\right)-{\mathrm{log}}_{b}\left(y\right)={\mathrm{log}}_{b}\left(\frac{x}{y}\right)$
• ${\mathrm{log}}_{b}\left({x}^{y}\right)=y{\mathrm{log}}_{b}\left(x\right)$
• ${\mathrm{log}}_{b}\left(\sqrt[y]{x}\right)=\frac{{\mathrm{log}}_{b}\left(x\right)}{y}$

## Change of Base

To find the logarithm in a base not supported on a calculator, b, the following formula can be used.

${\mathrm{log}}_{b}\left(x\right)=\frac{{\mathrm{log}}_{y}\left(x\right)}{{\mathrm{log}}_{y}\left(b\right)}$

Such that

${\mathrm{log}}_{3}\left(81\right)=\frac{{\mathrm{log}}_{10}\left(81\right)}{{\mathrm{log}}_{10}\left(3\right)}=4$