How do you evaluate log_5 (5^-1)?

Oct 29, 2016

Use the rule $\log {a}^{n} = n \log a$.

$= - 1 {\log}_{5} \left(5\right)$

Use the change of base rule ${\log}_{a} \left(n\right) = \log \frac{n}{\log} a$.

$= - 1 \log \frac{5}{\log} 5$

$= - 1$

Hopefully this helps!

Oct 29, 2016

${\log}_{5} \left({5}^{- 1}\right)$

$= - {\log}_{5} \left(5\right)$

$= - 1$

This is because:

${\log}_{a} \left({m}^{n}\right) = x$

Which means that:

${a}^{x} = {m}^{n}$

Then:

${a}^{\frac{x}{n}} = m$

As a result:

${\log}_{a} \left(m\right) = \frac{x}{n}$

And this implies that:

$x = n \cdot {\log}_{a} \left(m\right)$

Now, considering the fact that $x = {\log}_{a} \left({m}^{n}\right)$ as stated at the beginning of this proof, we can say that:

$n \cdot {\log}_{a} \left(m\right) = {\log}_{a} \left({m}^{n}\right)$

And this is why ${\log}_{5} \left({5}^{-} 1\right)$ evaluated is equal to $- 1$.