# Proving Logarithmic identity

Today, I read these two new Logarithmic identity $\displaystyle$ $$a^{\log_a m} = m$$ $$\log_{a^q}{m^p} = \frac{p}{q} \log_a m$$ Both of them seems new to me,so even after solving some problems (directly) based on thesm I haven't fully understood how they holds good,Could anybody show me how to prove them ?

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You know that $y = \log_b x \iff b^y = x$, right? – Djaian Nov 18 '10 at 9:57
To restate Djaian's hint: the logarithm and exponential are inverses of each other. – J. M. Nov 18 '10 at 10:03
Ok.That is for the first one I got it now ;) how about the second one ? – Quixotic Nov 18 '10 at 10:06
For the second identity: you're aware of the "change-of-base" formula, yes? – J. M. Nov 18 '10 at 10:07
Edit to $a^{\log_am}=m$ – Américo Tavares Nov 18 '10 at 10:08

1. This is a statement of the fact that the functions $a^x$ and $\log_a\;x$ are inverses of each other; thus, $\log_a\;a^x=a^{\log_a\;x}=x$
$\log_{a^q}\;m^p=\frac{\log_c\;m^p}{\log_c\;a^q}=\frac{p\;\log_c\;m}{q\;\log_c\;a}=\frac{p}{q}\log_a\;m$