Why does $(\log_x y)(\log_y x) = 1$?

Question

I just noticed that the product of two different logs with the bases switched around will always equal one. $(\log_x y)(\log_y x) = 1$

Why is this the case? What is the algebraic proof?

Thanks!

Answer 1

If you define logarithm as $$ \log_x y = \frac{\log y}{\log x} $$ where $\log$ is the natural logarithm (The inverse of $\exp$) Then $$ (\log_x y) (\log_y x) = \frac{\log y}{\log x} \frac{\log x}{\log y} = 1 $$

Answer 2

Note: $log_a(b) = \frac{log(b)}{log(a)}$. Therefore, $(\log_x y)(\log_y x) = 1 \implies \frac{\log{x}}{\log{y}} * \frac{\log{y}}{\log{x}} = 1 \implies 1 = 1$

Answer 3

Remember that $\log_x y$ is the value of an exponent $n$ that, when $x$ is raised to the $n^{\rm th}$ power, gives $y$: that is to say, $$\log_x y = n \quad \iff \quad x^n = y.$$ Therefore, if $\log_x y = n$ and $\log_y x = m$, we have $$x^n = y, \quad y^m = x,$$ and it becomes immediately obvious that $$x^{mn} = (x^n)^m = y^m = x,$$ or $mn = 1$, and the result follows.

Answer 4

$$(\log_x y)(\log_y x)=\log_x y^{(\log_y x)}=\log_xx=1$$

using:

1. $(\log_ab)c=\log_a(b^c)$
2. $d^{\log_df}=f$
3. $\log_gg=1$