# Thread: combining/simplifying logs (part 2)

1. ## combining/simplifying logs (part 2)

Here is the 2nd one which asks to simplify the whole thing as a logarithm. Again, not sure how to take the integers at the end and convert them into log form. Thanks.

2. Originally Posted by Evan.Kimia

Here is the 2nd one which asks to simplify the whole thing as a logarithm. Again, not sure how to take the integers at the end and convert them into log form. Thanks.
$\displaystyle 3 = log_4 (4^3)$

edit:

before that I would change $\displaystyle 2log_2(8) = 2log_2(2^3) = 6$.

Simplify the two integers and then make that into a log with base 4 to match that of x and y

3. Thank you, but im a bit confused how to go about changing the base, or if i even have to to log base 4.

ps., the correct answer is log base 4 (xy^2 over 64)

4. We know that $\displaystyle log_a(a) = 1$

Note that $\displaystyle 2log_2(8) = 6$ because$\displaystyle 8=2^3$

$\displaystyle 2log_4(y) = log_4(y^2)$

$\displaystyle -6+3 = -3$

As $\displaystyle -3 = log_4(4^{-3})$ (see the rule above)

So overall we can rewrite it as $\displaystyle log_4(x)+log_4(y^2)+log_4 \left(\frac{1}{64}\right)$

You can then combine the logs to give:

$\displaystyle log_4 \left(\frac{xy^2}{64}\right)$

5. ah! thank you!