Logarithm⇓⇓⇓⇓⇓

Question

Logarithm⇓⇓⇓⇓⇓

anonymous · Answer

$$\frac{ 1 }{ 3 }\log_{7}64+\frac{ 1 }{ 2 }\log_{7}121=\log_{7}x   $$

tkhunny · Answer

$64 = 4^{3}$

$121 = 11^{2}$

anonymous · Answer

whats the next step?

anonymous · Answer

where do in plug in that info

tkhunny · Answer

What did you do with that information?

You have logarithm rules.  $a\log(b) = \log\left(b^{a}\right)$

anonymous · Answer

I don't get it...

anonymous · Answer

where do i plug in $$4^3$$

tkhunny · Answer

What have you tried?  Show your work.  Just take a shot at it.  You can't break it.

anonymous · Answer

I don't understand what you are saying with $$4^3$$ is 3 a and 4 b if you tell me this i can figure it out

tkhunny · Answer

Why do you wonder about ONLY $4^{3}$?  I said, $4^{3} = 64$.  This answers all your questions, so far.

anonymous · Answer

Because $$4^3 and 11^2$$ should have the same labeling

tkhunny · Answer

That doesn't mean anything.

$\dfrac{1}{3}\log(64) = \log\left(64^{(1/3)}\right)$

anonymous · Answer

I know what to do if you help me with the labeling

tkhunny · Answer

It would help if I knew what that meant.

$(1/3)\log(64) = (1/3)\log\left(4^{3}\right) = (1/3)(3)\log(4) = log(4)$

$(1/2)\log(121) = (1/3)\log\left(11^{2}\right) = (1/2)(2)\log(11) = log(11)$

anonymous · Answer

I needed to know if in $$\log(b ^{a}) $$ 4was b an 3 was a and you just proved above that it was thank you