I would appreciate some help with Logarithims.

Question

jojokiw3 · Answer

$$4^{\log_{4} 6-3 \log_{4}2+2 \log_{4}\sqrt{5}-4\log_{4}1}$$

jojokiw3 · Answer

From what I figured out, the answer is $$\frac{ 15 }{ 4 }$$ But I'd like to understand the process to this answer.

zzr0ck3r · Answer

how did you figure it out?

zzr0ck3r · Answer

show us what you did, that would be the best way for you to learn.

jojokiw3 · Answer

Well. Okay. so I guess $$4^{\log_{4}6-\log_{4}2^3+\log_{4}\sqrt{5}^{2}-\log_{4}1^{4}}$$

jojokiw3 · Answer

That's what I converted it to, and I'm not sure if I'm correct.

jojokiw3 · Answer

@zzr0ck3r

anonymous · Answer

$$4^{\log_{4}6-3\log_{4} 2+2\log_{4}5^{\frac{ 1 }{ 2 }} -4\log_{4} 1 }$$
$$=4^{\log_{4}6 }-\log_{4} 2^3+\log_{4} 5^{(\frac{ 1 }{ 2 })^2}-\log_{4} 1^4$$
you are going on right track

zepdrix · Answer

Hmm this is a cool problem :)
You have to apply a bunch of your log AND exponent rules

jojokiw3 · Answer

Okay, so then $$4^{\log_{4}\frac{ 30 }{ 8 }}$$

jojokiw3 · Answer

Then finally it's $$\frac{ 15 }{ 4 }$$

zepdrix · Answer

The exponential base 4, and log base 4, are inverses of one another.
Taking their composition will "undo" each other.
So yes you're left with the 30/8, and you simplify.

Yayyy good job \c:/

jojokiw3 · Answer

Haha thanks. I wish I had 3 medals to give out. -.-

zepdrix · Answer

hah no big deal XD let's give you one though, you did most of the work here

jojokiw3 · Answer

:P Haha thanks! Oh well. one more step closer to finishing pre-calc!