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

Question

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.

Answer 1

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. $\square$

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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.

Answer 2

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*}

Answer 3

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