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 ?
Proving Logarithmic identity
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$\begingroup$
algebra-precalculus
logarithms
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1You know that $y = \log_b x \iff b^y = x$, right? – 2010-11-18
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1To restate Djaian's hint: the logarithm and exponential are inverses of each other. – 2010-11-18
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0Ok.That is for the first one I got it now ;) how about the second one ? – 2010-11-18
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0For the second identity: you're aware of the "change-of-base" formula, yes? – 2010-11-18
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0Edit to $a^{\log_am}=m$ – 2010-11-18
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0@JM:Do you mean $\log_a m = \log_b m \cdot log_a b$ where b is any real $\gt 1$? – 2010-11-18
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0Okay I got it ! thanks J.M :) – 2010-11-18
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0I understood you meant the alternate form of it ... that is the division with base `e`. – 2010-11-18
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0Not really: the thing is that you can change one logarithm into another, and that's regardless of what your starting and ending bases you have. – 2010-11-18
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0Hm.that's the rule. – 2010-11-18
1 Answers
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Just so this doesn't remain unanswered:
This is a statement of the fact that the functions $a^x$ and $\log_a\;x$ are inverses of each other; thus, $\log_a\;a^x=a^{\log_a\;x}=x$
Letting c be a positive real number not equal to one:
$\log_{a^q}\;m^p=\frac{\log_c\;m^p}{\log_c\;a^q}=\frac{p\;\log_c\;m}{q\;\log_c\;a}=\frac{p}{q}\log_a\;m$