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I've encountered this quite a bit.

If I have ${\log(a)\over \log(b)} = c$ where $b$ is a known positive integer, what can be said about $a$ if $c$ needs to be an integer?

Winther
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user257765
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1 Answers1

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Then $\ln a = c\ln b$, and taking the exponential on both side you get that $a= b^c$. That is, $a$ is a power of $b$.

Clement C.
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  • Can $a,b$ and/or $c$ be negative? – user257765 Aug 08 '15 at 17:32
  • The logarithm is only defined on $(0,\infty)$, so for the equation to make sense both $a$ and $b$ have to be positive (and actually, $b > 1$, since otherwise you divide by zero). This implies $c$ has to be greater or equal to zero has well. – Clement C. Aug 08 '15 at 17:35
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    $\frac{\log(2^{-2})}{log(2)}=-2$ In this example, $c$ is negative. – randomgirl Aug 08 '15 at 18:05
  • @randomgirl sorry, yes -- I was confused and assumed in the comment that $a$ had to be an integer. Without this (which is, I guess, what the OP indeed had in mind), $c$ can be a negative integer (the only constraints are $a > 0$, and $b > 1$. – Clement C. Aug 08 '15 at 18:08