# How do you solve log_(1/7)x=-1?

Mar 16, 2018

I tried this:

#### Explanation:

We can use the definition of log:

${\log}_{b} a = x$

so that:

$a = {b}^{x}$

In our case:
${\log}_{\frac{1}{7}} x = - 1$
so that:
$x = {\left(\frac{1}{7}\right)}^{-} 1 = 7$
using a property of exponents.