# Prove? log_x(y)xxlog_y(x)=1

Jan 21, 2018

See explanation.

#### Explanation:

The change of base formula says ${\log}_{a} \left(b\right) = \log \frac{b}{\log} \left(a\right)$.

Applying that here we have:

${\log}_{x} \left(y\right) \cdot {\log}_{y} \left(b\right) = \log \frac{y}{\log} \left(x\right) \setminus \cdot \log \frac{x}{\log} \left(y\right)$

$= \log \frac{y}{\log} \left(y\right) \setminus \cdot \log \frac{x}{\log} \left(x\right) = 1$

Jan 22, 2018

By definition of the logarithm we have:

${\log}_{a} b = c \iff {a}^{c} = b$

If we apply the definition to ${\log}_{x} y$; then:

${\log}_{x} y = A \iff {x}^{A} = y$

And if we apply the definition to ${\log}_{y} x$; then

${\log}_{y} x = B \iff {y}^{B} = x$

And if we replace $x$ from the second expression into $x$ from the first expression; then we gte:

${\left({y}^{B}\right)}^{A} = y \iff {y}^{A B} = y \iff A B = 1$

And the result follows, as:

$A B = 1 \iff {\log}_{x} y \setminus {\log}_{y} x = B = 1$ QED

See below

#### Explanation:

There are a couple of ways to do this:

• Use the log base switch rule: ${\log}_{a} \left(b\right) = \frac{1}{\log} _ b \left(a\right)$

${\log}_{x} \left(y\right) \times {\log}_{y} \left(x\right) = 1$

$\frac{1}{\log} _ y \left(x\right) \times {\log}_{y} \left(x\right) = {\log}_{y} \frac{x}{\log} _ y \left(x\right) = 1$

• Use the log base change rule: ${\log}_{a} \left(b\right) = {\log}_{x} \frac{b}{\log} _ x \left(a\right)$

${\log}_{x} \left(y\right) \times {\log}_{y} \left(x\right) = 1$

$\log \frac{y}{\log} x \times \log \frac{x}{\log} y = \frac{\log x \log y}{\log x \log y} = 1$