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I have this equation

$p = r^l$

I know $p$ and $r$ and I would like to solve for $l$, how can I rearrange this formula to achieve this?
I initially came up with:

$l = \sqrt[r]{p}$

( $l$ is the $r$th root of $p$) but my intuition says this is wrong and after putting in numbers - I know this is wrong - how can I rearrange to solve for $l$?

the_velour_fog
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1 Answers1

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This is what the logarithm is for. We say $l=\log_r(p)$.

Fimpellizzeri
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  • thanks, is there an intermediate step between my function and your answer (Im hoping for an explanation of how you arrived at your solution) or are logarithms just something that you just have to know should be used whenever you see a certain pattern? – the_velour_fog Nov 21 '16 at 04:52
  • after putting in numbers, this seems correct. I will accept when min-accept-wait-time has elapsed, thanks :) – the_velour_fog Nov 21 '16 at 04:56
  • Well, there may be an intermediate step in the sense that you might have to learn about them. Basically, it boils down to observing that for $a>0$ the function $f(x)=a^x$ is injective for $x>0$, so it can be inverted. The inverse of this function is $g(x)=\log_a(x)$. Properties of the exponential get translated into properties of logarithms; for instance, $\log(xy)=\log(x)+\log(y)$. But aside from your knowledge of the existence of logarithms (and why they're allowed to exist), there is no intermediate step. – Fimpellizzeri Nov 21 '16 at 04:56
  • ah thanks, even though there is no intermediate step, I think I see why there is no single obvious intermediate step. It seems its because $p$ is captured into the inversion. in other words - a logarithm is a transform involving multiple operands not a single operand like most algebraic operations. In this case $r$ becomes the base and $p$ - simultaneously becomes the argument to the log function. – the_velour_fog Nov 21 '16 at 05:12