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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

2 Answers 2

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

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    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}$2012-11-15
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this evaluation is based on following basic rules

$\log_c ab=\log_ca+\log_cb$

$\log_c a^n=n\log_ca$

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