Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Let, $\frac{x(y+z-x)}{logx}=\frac{y(x+z-y)}{logy}=\frac{z(x+y-z)}{logz}=k$.

$logx=\frac{x(y+z-x)}{k}$, multiplying by $y$ on both sides,

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

Also, multiplying both sides by $z$ we get- $z\log x=\frac{zx(y+z-x)}{k}$ ........(2)

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

multiplying both sides by $x$, we get-

$x\log y=\frac{xy(x+z-y)}{k}$ .......(3)

Also, multiplying both sides by $z$ we get- $z\log y=\frac{zy(x+z-y)}{k}$ ...........(4)

And, $logz=\frac{z(x+y-z)}{k}$, multiplying both sides by $y$, we get- $y\log z=\frac{yz(x+y-z)}{k}$ .......(5)

Also, multiplying both sides by $x$ we get- $xlogz=\frac{xz(x+y-z)}{k}$ ..........(6)

Adding $1$ and $3$, we have- $\log \left(x^y\right)\left(y^x\right)=\frac{xyz}{k}$

Adding $4$ and $5$, we have- $\log \left(y^z\right)\left(z^y\right)=\frac{xyz}{k}$

Adding $2$ and $6$, we have- $\log \left(z^x\right)\left(x^z\right)=\frac{xyz}{k}$

Comparing the $RHS$ of above and removing $]\log$, we have-

$\left(x^y\right)\left(y^x\right)=\left(y^z\right)\left(z^y\right)=\left(x^z\right)\left(z^x\right)$

Hence proved.