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I'm trying to figure out how to go about proving this statement:

$$n^{\log(a)} = a^{\log(n)}$$

I'm told that I cannot prove from both sides. I tried to $\log$ the first side to get:

$\log(n)\log(a)$

But I'm not sure where to go from here, any ideas would be much appreciated.

user3739406
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  • what do you get if you take the log of the other side? – lulu Oct 14 '15 at 20:39
  • Thing is I can't do that, I can only work from one side. (Proove left side = Right Side) – user3739406 Oct 14 '15 at 20:40
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    Ok. rewrite $log(n)log(a)$ as $log(a)log(n)$ and exponentiate. – lulu Oct 14 '15 at 20:40
  • I'm quite confused now, am I even allowed to log the first side? Doesn't that mean I would have to Log the second side as well, but my professor tells me I can only work from one side. – user3739406 Oct 14 '15 at 20:45
  • My (most recent) suggestion only works with one side. We know that $xlog(y)=log(x^y)$ so $log(a)log(n) = log(n^{log(a)})$ and $log(n)log(a)=log(a^{log(n)})$. But of course $log(a)log(n)=log(n)log(a)$. – lulu Oct 14 '15 at 20:51
  • Forgive me for trying to read your professor's mind, but I imagine what she means is that you cannot start with the assumption that $A=B$, manipulate both sides simultaneously, and then arrive at a true statement. However, it is perfectly fine to take the log of both sides, $\log(A)$ and $\log(B)$, separately, and then observe that they are equal. – Jeff Oct 14 '15 at 21:01

3 Answers3

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I'm not sure if you can use it, but it is well known that

$$x^y = e^{(\ln x)y} $$

where $\ln$ is the natural logarithm and $e$ is Eulers constant. Just apply this.

Stefan Hante
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  • I'm not sure I follow, care to explain, I'm not familiar with this property? – user3739406 Oct 14 '15 at 20:42
  • Do you know that $y\ln x = \ln \left(x^y\right)$? – Chinny84 Oct 14 '15 at 20:46
  • Since the exponential function $x\mapsto e^x$ and the natural logarithm are inverse it is $e^{\ln x} = x$. Further it is $(e^{\ln x})^y = e^{\ln y}$. These two are in every mathematical table. Knowing this you can use $x=n$ and $y=\ln a$. Then apply what I wrote, switch the order of the logarithms by commutativity and apply again what I wrote. – Stefan Hante Oct 14 '15 at 20:46
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We can just do some simple manipulation to get the result. We can take a log and exponentiate the left side:

$$n^{\log(a)}=e^{\log(n^{\log(a)})}=e^{\log(a)\log(n)}=e^{\log(n)\log(a)}=e^{\log(a^{\log(n)})}=a^{\log(n)}.$$

superckl
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Suppose $n=a^x$, then $\log n=x\log a$ by taking logs.

Then, by substitution, $n^{\log a}=(a^x)^{\log a}=\dots$

How to find the method - well you start with $n$ to some power, and you want to end with $a$ to some power, so it is natural to express $n$ as a power of $a$, and once this is done the result drops out.

Mark Bennet
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