Question

$\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(z+x-y)}{\log z},$ then prove that $x^y{y}^x=z^y{y}^z=x^z{z}^x$.

Answer 1

We have,

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

Let,

$\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(x+y-z)}{\log z}=k$

Then,

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

$\log y=\frac{y(z+x-y)}{k}.......\left(2\right)$

$\mathrm{logz}=\frac{z(x+y-z)}{k}........\left(3\right)$

Multiplying by $$y$$in equation (1) and (3) to and multiplying by $x$ in equation (2) and we get,

$y\log x=\frac{xy(y+z-x)}{k}......\left(4\right)$

$x\log y=\frac{xy(z+x-y)}{k}.......\left(5\right)$

$y\mathrm{logz}=\frac{yz(x+y-z)}{k}........\left(6\right)$

Multiplying by $$z$$in equation (1) and (2) to and multiplying by $x$ in equation (3) and we get,

$z\log x=\frac{xz(y+z-x)}{k}......\left(7\right)$

$z\log y=\frac{yz(z+x-y)}{k}.......\left(8\right)$

$x\mathrm{logz}=\frac{xz(x+y-z)}{k}........\left(9\right)$

On adding equation (4) and (5),(6) and (8),and (7) and (9) to respectively and we get,

$y\log x+x\log y=\frac{xy(y+z-x)}{k}+\frac{xy(z+x-y)}{k}$

$\log x^y+\log y^x=\frac{x{y}^2+xyz-x^{2}y+xyz+x^{2}y-x{y}^2}{k}$

$\log \left(x^y{y}^x\right)=\frac{2xyz}{k}.......\left(a\right)$

$y\mathrm{logz}+z\log y=\frac{yz(x+y-z)}{k}+\frac{yz(z+x-y)}{k}$

$\log \left(z^y{y}^z\right)=\frac{2xyz}{k}.......\left(b\right)$

$z\mathrm{logx}+x\mathrm{logz}=\frac{xz(y+z-x)}{k}+\frac{xz(x+y-z)}{k}$

$\log \left(x^z{z}^x\right)=\frac{2xyz}{k}.......\left(c\right)$

Now, equation (a),(b) and (c) to,

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

Then, $x^y{y}^x=z^y{y}^z=x^z{z}^x$

Hence, this is the answer.