# combining/simplifying logs (part 2)

• Oct 6th 2009, 10:48 AM
Evan.Kimia
combining/simplifying logs (part 2)
http://img25.imageshack.us/img25/2971/log2vh.jpg

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. (Headbang) Thanks.
• Oct 6th 2009, 10:58 AM
e^(i*pi)
Quote:

Originally Posted by Evan.Kimia
http://img25.imageshack.us/img25/2971/log2vh.jpg

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. (Headbang) 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
• Oct 6th 2009, 11:29 AM
Evan.Kimia
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)
• Oct 6th 2009, 11:44 AM
e^(i*pi)
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:

$\displaystyle log_4 \left(\frac{xy^2}{64}\right)$
• Oct 6th 2009, 11:56 AM
Evan.Kimia
ah! thank you!