# How do you evaluate Log_sqrt3 243 ?

Mar 13, 2016

$10$

#### Explanation:

We should try to write $243$ as a power of $3$.

An example of the method we should attempt can be shown more easily in evaluating ${\log}_{2} 8$:

${\log}_{2} 8 = {\log}_{2} {2}^{3} = 3 {\log}_{2} 2 = 3$

The first step we should take for ${\log}_{\sqrt{3}} 243$ is to recognize that $243 = {3}^{5}$.

${\log}_{\sqrt{3}} 243 = {\log}_{\sqrt{3}} {3}^{5}$

But we still haven't addressed the issue that we want a base of $\sqrt{3}$, not $3$.

We should use the fact that ${\left(\sqrt{3}\right)}^{2} = 3$, like so:

${\log}_{\sqrt{3}} {3}^{5} = {\log}_{\sqrt{3}} {\left({\left(\sqrt{3}\right)}^{2}\right)}^{5}$

Using the rule that ${\left({x}^{a}\right)}^{b} = {x}^{a b}$, we multiply $2$ and $5$ to see that

${\log}_{\sqrt{3}} {\left({\left(\sqrt{3}\right)}^{2}\right)}^{5} = {\log}_{\sqrt{3}} {\left(\sqrt{3}\right)}^{10}$

Now, simplify as was done earlier.

${\log}_{\sqrt{3}} {\left(\sqrt{3}\right)}^{10} = 10 {\log}_{\sqrt{3}} \sqrt{3} = 10$