# Prove that x^((logy-logz))y^((logz-logx))z^((logx-logy))=1?

Mar 20, 2017

#### Explanation:

Prove:

${x}^{{\log}_{10} \left(y\right) - {\log}_{10} \left(z\right)} {y}^{{\log}_{10} \left(z\right) - {\log}_{10} \left(x\right)} {z}^{{\log}_{10} \left(x\right) - {\log}_{10} \left(y\right)} = 1$

Use the base 10 logarithm on both sides:

${\log}_{10} \left({x}^{{\log}_{10} \left(y\right) - {\log}_{10} \left(z\right)} {y}^{{\log}_{10} \left(z\right) - {\log}_{10} \left(x\right)} {z}^{{\log}_{10} \left(x\right) - {\log}_{10} \left(y\right)}\right) = {\log}_{10} \left(1\right)$

The right side becomes 0:

${\log}_{10} \left({x}^{{\log}_{10} \left(y\right) - {\log}_{10} \left(z\right)}\right) + {\log}_{10} \left({y}^{{\log}_{10} \left(z\right) - {\log}_{10} \left(x\right)}\right) + {\log}_{10} \left({z}^{{\log}_{10} \left(x\right) - {\log}_{10} \left(y\right)}\right) = 0$

Use the property of logarithms ${\log}_{b} \left({a}^{c - d}\right) = {\log}_{b} \left(a\right) \left(c - d\right)$

${\log}_{10} \left(x\right) \left({\log}_{10} \left(y\right) - {\log}_{10} \left(z\right)\right) + {\log}_{10} \left(y\right) \left({\log}_{10} \left(z\right) - {\log}_{10} \left(x\right)\right) + {\log}_{10} \left(z\right) \left({\log}_{10} \left(x\right) - {\log}_{10} \left(y\right)\right) = 0$

Use the distributive property on all of the parenthesis:

log_10(x)log_10(y)-log_10(x)log_10(z)+log_10(y)log_10(z)-log_10(y)log_10(x)+log_10(z)log_10(x)-log_10(z)log_10(y))=0

Begin canceling terms:

cancel(log_10(x)log_10(y))-log_10(x)log_10(z)+log_10(y)log_10(z)cancel(-log_10(y)log_10(x))+log_10(z)log_10(x)-log_10(z)log_10(y))=0

cancel(-log_10(x)log_10(z))+log_10(y)log_10(z)cancel(+log_10(z)log_10(x))-log_10(z)log_10(y))=0

$\cancel{+ {\log}_{10} \left(y\right) {\log}_{10} \left(z\right)} \cancel{- {\log}_{10} \left(z\right) {\log}_{10} \left(y\right)} = 0$

$0 = 0$

Q.E.D.

Mar 20, 2017

Please see below for the proof.

#### Explanation:

We will not be writing base as $10$, hence $\log p = {\log}_{10} p$

Now let ${x}^{a} = {y}^{b}$, then $a \log x = b \log y$

and $b = a \times \log \frac{x}{\log} y$

Hence ${x}^{\log y - \log z} = {y}^{\left(\log y - \log z\right) \times \log \frac{x}{\log} y}$

and ${z}^{\log x - \log y} = {y}^{\left(\log x - \log y\right) \times \log \frac{z}{\log} y}$

and hence

${x}^{\left(\log y - \log z\right)} {y}^{\left(\log z - \log x\right)} {z}^{\left(\log x - \log y\right)}$

= ${y}^{\left(\left(\log y - \log z\right) \times \log \frac{x}{\log} y\right)} {y}^{\left(\log z - \log x\right)} {y}^{\left(\left(\log x - \log y\right) \times \log \frac{z}{\log} y\right)}$

= ${y}^{\left(\left(\log y - \log z\right) \times \log \frac{x}{\log} y\right) + \left(\log z - \log x\right) + \left(\left(\left(\log x - \log y\right) \times \log \frac{z}{\log} y\right)\right)}$

= ${y}^{\log x - \frac{\log z \log x}{\log} y + \log z - \log x + \frac{\log x \log z}{\log} y - \log z}$

= ${y}^{0} = 1$

Mar 20, 2017

See below.

#### Explanation:

${x}^{\log} \left(\frac{y}{z}\right) {y}^{\log} \left(\frac{z}{x}\right) {z}^{\log} \left(\frac{y}{x}\right) = 1$

or

$\log \left(\frac{y}{z}\right) \log \left(x\right) + \log \left(\frac{z}{x}\right) \log \left(y\right) + \log \left(\frac{y}{x}\right) \log \left(z\right) = 0$

or

${\left(\frac{y}{z}\right)}^{\log} \left(x\right) {\left(\frac{z}{x}\right)}^{\log} \left(y\right) {\left(\frac{x}{y}\right)}^{\log} \left(z\right) = 1$

or

${x}^{\log} \left(\frac{z}{y}\right) {y}^{\log} \left(\frac{x}{z}\right) {z}^{\log} \left(\frac{y}{x}\right) = 1 = {x}^{\log} \left(\frac{y}{z}\right) {y}^{\log} \left(\frac{z}{x}\right) {z}^{\log} \left(\frac{y}{x}\right) = {x}^{-} \log \left(\frac{z}{y}\right) {y}^{-} \log \left(\frac{x}{z}\right) {z}^{-} \log \left(\frac{y}{x}\right)$

then

x^log(z/y)y^log(x/z) z^log(y/x)=1/(x^log(z/y)y^log(x/z) z^log(y/x)

so

${x}^{\log} \left(\frac{z}{y}\right) {y}^{\log} \left(\frac{x}{z}\right) {z}^{\log} \left(\frac{y}{x}\right) = 1$ is an identity.