How do you solve $(1/5)^x=10$ ?

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Answer 1 · 2015-09-10T04:15:56

$x=log10/(log1-log5)$

By taking logarithms in both sides we get

$log(1/5)^x=log10=>x(log1-log5)=log10=>x=log10/(log1-log5)$

Answer 2 · 2015-09-10T12:02:44

Use properties of exponents and logs to reformulate and solve, finding:

$x = -1/log(5)$

$(1/5)^x = 5^(-x)$

So $log((1/5)^x) = log(5^(-x)) = -x*log(5)$

$log(10) = 1$

So taking logs of both sides, the original equation becomes:

$-x*log(5) = 1$

Divide both sides by $-log(5)$ to get:

$x = -1/log(5)$

How do you solve (1/5)^x=10(1/5)^x=10?