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.

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- Oct 6th 2009, 10:48 AMEvan.Kimiacombining/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 AMe^(i*pi)
- Oct 6th 2009, 11:29 AMEvan.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 AMe^(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 AMEvan.Kimia
ah! thank you!