# combining/simplifying logs (part 2)

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• 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.

$3 = log_4 (4^3)$

edit:

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

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 $log_a(a) = 1$

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

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

$-6+3 = -3$

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

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

You can then combine the logs to give:

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