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If $x>0$ and $x^a=x^b$ can one assume that $a=b$? The answer says it's not right. I've tried coming up with a counter-example but keep failing.

Thanks in advance!

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    $\begingroup$ No, $1^0=1^1=1$, but $0\neq 1$. $\endgroup$ – Dietrich Burde May 16 '16 at 19:15
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Let's take a log and see what happens:

$$\log(x^a) = \log(x^b)\Longrightarrow a\log x = b\log x \Longrightarrow (b-a)\log x = 0.$$

There are two cases to consider here. I'll let you figure out what those are.

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