## Inverted Logarithm

#### Problem

Prove that $\log_{a}(x) \log_{b}(y) = \log_{b}(x) \log_{a}(y)$.

#### Solution

We shall begin by proving two fundamental properties of logarithms:

1. $\log_{a}(x) = \dfrac{\log_{b}(x)}{\log_{b}(a)}$
2. $\log_{b}(a) = \dfrac{1}{\log_{a}(b)}$

Clearly the first result is well known, as it is the method often used by students to evaluate logarithms in different bases on calculators; for example, $\log_{2}(32) = \dfrac{\log_{10}(32)}{\log_{10}(2)} = 5$. The second result is perhaps less familiar and is called the reciprocal property of logarithms.

1. Let $log_{a}(x) = c \implies x = a^c$ (from definition of logarithms).

$\therefore \log_{b}(x) = \log_{b}(a^c) = c \log_{b}(a)$

$\therefore c = \log_{a}(x) = \dfrac{\log_{b}(x)}{\log_{b}(a)}$ [1]

2. Let $\log_{a}(b) = c \implies b = a^c$.

$\therefore b^{1/c} = a \implies \dfrac{1}{c} = \log_{b}(a)$

$\therefore \log_{b}(a) = \dfrac{1}{c} = \dfrac{1}{\log_{a}(b)}$ [2]

Now we shall prove that $\log_{a}(x) \log_{b}(y) = \log_{b}(x) \log_{a}(y)$.

Using [1], $\log_{a}(x) = \dfrac{\log_{b}(x)}{\log_{b}(a)}$ and $\log_{b}(y) = \dfrac{\log_{a}(y)}{\log_{a}(b)}$.

$\therefore \log_{a}(x) \log_{b}(y) = \dfrac{\log_{b}(x) \log_{a}(y)}{\log_{b}(a) \log_{a}(b)}$

Using [2], $\log_{b}(a) = \dfrac{1}{\log_{a}(b)} \implies \log_{b}(a) \log_{a}(b) = 1$.

Hence $\log_{a}(x) \log_{b}(y) = \log_{b}(x) \log_{a}(y)$   Q. E. D.

Problem ID: 315 (18 Mar 2007)     Difficulty: 3 Star

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