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My question is in the title. Why is:

$$\frac{\log(y)}{\log(x)} = \log_{x}(y)$$

I realize this was something I learned perhaps in middle school, but until my interest in more analytical math, I had never thought about the proof for it.

SundayLi
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user8393252
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3 Answers3

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This is a great question! I'll assume $\log$ is the natural logarithm (but any other base will work, of course).

We want to prove that $\frac{\log(y)}{\log(x)} = \log_x(y)$.

$\log_x(y)$ means "the power to which $x$ must be raised to produce $y$", so in other words we need to prove that $$x^{\log(y)/\log(x)} = y.$$

Let's do it! We know that $x = e^{\log(x)}$, so we compute:

$$x^{\log(y)/\log(x)} = (e^{\log(x)})^{\log(y)/\log(x)} = e^{\log(y)} = y,$$

as desired. $\square$

For the pedants in the crowd: of course I did not start by giving a formal definition of $\log_x$, $e$, etc. However, this is clearly not the point of the question.

diracdeltafunk
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  • The definition of $\log$ is not pedantry: $\log_x(y)=\frac{\log(y)}{\log(x)}$ could litterally be a definition. Actually, your answer defines $\log_x$! (As the inverse function of $t\mapsto x^t$) – Taladris Mar 31 '23 at 01:13
  • @Taladris of course you're right! But I believe that OP is looking for motivation for this fact, not necessarily a proof. If this equality is definitional, then the correct proof is 1-line long, but the motivation behind the scenes is still something like what I wrote. My answer is phrased as a proof, but is really more a heuristic justification, saying "if you want arithmetic to have so-and-so usual properties, you need to also accept that $\log_x(y) = \log(x)/\log(y)$". – diracdeltafunk Mar 31 '23 at 01:16
  • The point of my suggested approach was to avoid (the scarier) $x^b = e^{b\log x}$. – Ted Shifrin Mar 31 '23 at 01:39
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Another perspective on @diracdeltafunk's answer. By definition we have that $\log_x(y)$ is the solution (for $t$) of $x^t=y$. But we solve this exponential equation the way we solve any exponential equation; by taking logs (with respect to your favorite base): \begin{align*} x^t &= y \\ t \log(x) &= \log(y) \\ t &= \log(y)/\log(x). \end{align*}

Jamie Radcliffe
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Let $z=\log_xy$. By the definition of logarithm, we can rewrite it as $$x^z=y$$ Then taking logarithm on both sides yields $$ \begin{aligned} & \log \left(x^z\right)=\log y \\ \Leftrightarrow \quad &z \log x=\log y \\ \Leftrightarrow \quad &z=\frac{\log y}{\log x} \\ \Leftrightarrow \quad &\log _x y=\frac{\log y}{\log x} \end{aligned} $$

Lai
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