`\log_{`

`base`}`number` = {?}

`answer`

If `\log_{b}x=y`

, then `b^y=x`

.

First, try to write

, the number we are taking the logarithm of, as a power of `number`

, the base of the logarithm.`base`

can be expressed as `number`

.`power_string`

can be expressed as `number`

.`base`^`answer`

, so `base`^`answer`=`number``\log_{`

.`base`}`number`=`answer`