Question

If $\frac{x(y+z-x)}{logx}=\frac{y(z+x-y)}{logy}=\frac{z(x+y-z)}{logz}$, prove that $x^y{y}^x=z^y{y}^z=x^z{z}^x$.

Answer 1

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

$\log x=k(xy+xz-x^2)\cdots \left(1\right)$

$\log y=k(yz+xy-y^2)\cdots \left(2\right)$

$\log z=k(zx+yz-z^2)\cdots \left(3\right)$

$\left(2\right)\times x⟹x\log y=k(xyz+x^{2}y-x{y}^2)\cdots \left(4\right)$

$\left(3\right)\times y⟹y\log x=k(x{y}^2+xyz-x^{2}y)\cdots \left(5\right)$

$\left(4\right)+\left(5\right)⟹y\log x+x\log y=2k\left(xyz\right)⟹\log x^y{y}^x=2k\left(xyz\right)$

Similarly $\log z^y{y}^z=2k\left(xyz\right),\log x^z{z}^x=2k\left(xyz\right)$

$\log x^y{y}^x=\log z^y{y}^z=\log x^z{z}^x⟹x^y{y}^x=z^y{y}^z=x^z{z}^x$