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today I'm in doubt on calculating the follow expression $\log_4 3 * \log_9 32$

Changing all to base 4: Working on: $\log_4 3 * \dfrac{\log_4 32}{\log_4 9}$

Ending with: $\log_4 3 * \dfrac{2 + \log_4 2}{2*\log_4 3}$

There's a way to simplify it more ? Also, do you know any resource explaining more on rules on every kind of operation with logs ?

Thanks in advance

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2 Answers 2

up vote 6 down vote accepted

$\log_4 2$ is $\frac{1}{2}$. So it evaluates to $\frac{5}{4}$.

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Thanks, I've evaluated it to $\log_4 3 * \dfrac{\frac{5}{4}}{2*\log_4 3}$ –  aajjbb Nov 15 '12 at 17:43
1  
No. The whole expression evaluates to... wait did you mean something else by chance? I mean could you put brackets on your actual problem so that we can read it correctly. I am wondering why u didnt cancel $\log_4 3$. Also the numerator is $\frac{5}{2}$. I thought you could cancel the $\log_4 3$ and you would get $\frac{5}{4}$. But apparently your question was something different... –  Gautam Shenoy Nov 15 '12 at 17:48
    
Sorry, you're right, I have forgotten to cancel the $\log_4 3$ from the expression, so the final result will be $\dfrac{5}{4}$ –  aajjbb Nov 15 '12 at 18:11

this evaluation is based on following basic rules

$\log_c ab=\log_ca+\log_cb$

$\log_c a^n=n\log_ca$

$\log_ab=\frac{\log_ca}{\log_cb}$

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