When a=3 & b=2, then loga b^a x log b a^b= ?

I'll put the question picture on the comment

Question

When a=3 & b=2, then loga b^a x log b a^b= ?

I'll put the question picture on the comment

anonymous · Answer

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mathmale · Answer

Simply substitute 3 for a and 2 for b.

anonymous · Answer

I know that it'll be log2 9 x log3 8, but I'm getting confused on how to do the rest. 
My teacher didn't teach me logarithm really well when I was in high school so I don't really understand.

Can you help me solve the rest?

anonymous · Answer

Okay we know that $$\log _{b} a = \frac{ \log a }{ \log b }$$

anonymous · Answer

that is, if we have to express in base 10

anonymous · Answer

then:
$$\frac{ \log 3^{2} \times \log 2^{3}}{\log 2 \times \log 3 }=\frac{ 2 \log 3 \times 3 \log 2}{\log 2 \times \log 3} $$
Eliminate log 3 & log 2 and the answer is 6?

anonymous · Answer

Yes @Jrc01. Had network issues over here...

anonymous · Answer

Ah, I got it now. No problem & thank you very much for the help!!

anonymous · Answer

Yw :)