# How do you evaluate log_sqrt7 49?

Apr 4, 2016

$4$

#### Explanation:

We should attempt to write $49$ as a power of $\sqrt{7}$.

First, simply rewrite $49$ as ${7}^{2}$.

${\log}_{\sqrt{7}} 49 = {\log}_{\sqrt{7}} \left({7}^{2}\right)$

This can be rewritten using the rule:

${\log}_{a} \left({b}^{c}\right) = c \cdot {\log}_{a} b$

Thus, we have

${\log}_{\sqrt{7}} \left({7}^{2}\right) = 2 {\log}_{\sqrt{7}} 7$

Now, we can write that $7 = {\left(\sqrt{7}\right)}^{2}$.

$2 {\log}_{\sqrt{7}} {\left(\sqrt{7}\right)}^{2}$

Rewrite using the rule we found earlier.

$2 \cdot 2 {\log}_{\sqrt{7}} \sqrt{7}$

Note that

${\log}_{a} a = 1$

So we are just left with

$2 \cdot 2 {\log}_{\sqrt{7}} \sqrt{7} = 4$

We can test this by using our rules of logarithms. We said that:

${\log}_{\sqrt{7}} 49 = 4$

This means that ${\left(\sqrt{7}\right)}^{4} = 49$

$\sqrt{7} \times \sqrt{7} \times \sqrt{7} \times \sqrt{7} = 7 \times 7 = 49$