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

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Answer 1 · 2018-03-16T18:36:49

I tried this:

We can use the definition of log:

$log_ba=x$

so that:

$a=b^x$

In our case:
$log_(1/7)x=-1$
so that:
$x=(1/7)^-1=7$
using a property of exponents.

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