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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!

mathtron999
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    It is a case of the change of base formula: $\log_x(y)=\frac{\log_y(y)}{\log_y(x)}$. The change of base formula follows fairly easily from the familiar rules of exponents; you can ask about that if you like. – Ian Dec 29 '16 at 19:56
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    If you want to prove this directly without the base change formula, you could go $$\log_x(y)\log_y(x) = \log_y(x^{\log_x(y)}) = \log_y(y) = 1.$$ – Mees de Vries Dec 29 '16 at 19:58

4 Answers4

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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.

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

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

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

using:

  1. $(\log_ab)c=\log_a(b^c)$
  2. $d^{\log_df}=f$
  3. $\log_gg=1$
πr8
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