# How do you solve log_7 x= -1?

May 11, 2016

$x = {7}^{- 1} = \frac{1}{7}$

#### Explanation:

The log function is asking us what power of the base gives us the argument of the $\log$, in equation form:

If $\text{ " x=b^y " }$ then $\text{ } {\log}_{b} \left(x\right) = y$

The inverse of a $\log$ function is a power of the base, i.e.

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

The way we use this is to apply the "inverse" to both sides of an equation. Starting with the equation from our question

${\log}_{7} x = - 1$

Raising both sides of the equation to the power of the base, $7$, we get

${7}^{{\log}_{7} x} = {7}^{- 1}$

Then simplifying both sides using our expression from above:

$x = {7}^{- 1} = \frac{1}{7}$