Question

If $\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(x+y-z)}{\log z}$, then $x^y{y}^x+z^y{y}^z-2x^z{z}^x$ is equal to

• A. $-1$
• B. $0$
• C. $2$
• D. not independent of $x,y,z$

Answer 1

The correct option is B $0$

Let $\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(x+y-z)}{\log z}=k$
$\frac{x(y+z-x)}{\log x}=k\Rightarrow \log x=\frac{x(y+z-x)}{k}$ .........................(A)
Multiplying both sides of (A) by $y$ we get
$y\log x=\frac{xy(y+z-x)}{k}\Rightarrow \log x^y=\frac{xy(y+z-x)}{k}$ ......................(1)
And multiplying both sides of (A) by z we get
$z\log x=\frac{xz(y+z-x)}{k}\Rightarrow \log x^z=\frac{xz(y+z-x)}{k}$ ..........................(2)
.
$\frac{y(z+x-y)}{\log y}=k\Rightarrow \log y=\frac{y(z+x-y)}{k}$ ...................(B)
Multiplying both sides of (B) by $x$ we get
$x\log y=\frac{xy(z+x-y)}{k}\Rightarrow \log y^x=\frac{xy(z+x-y)}{k}$ .........................(3)
And multiplying both sides of (B) by z we get
$z\log y=\frac{yz(z+x-y)}{k}\Rightarrow \log y^z=\frac{yz(z+x-y)}{k}$ ...............................(4)
$\frac{z(x+y-z)}{\log z}=k\Rightarrow \log z=\frac{z(x+y-z)}{k}$ ...............................(C)
Multiplying both sides of (C) by $x$ we get
$x\log z=\frac{xz(x+y-z)}{k}\Rightarrow \log z^x=\frac{xz(x+y-z)}{k}$ ..............................(5)
And multiplying both sides of (C) by y we get
$y\log z=\frac{yz(x+y-z)}{k}\Rightarrow \log z^y=\frac{xz(x+y-z)}{k}$ ...................................(6)
Adding (1) and (3)
$\log x^y{y}^x=\frac{xyz}{k}\Rightarrow x^y{y}^x=e^{2xyz/k}$
Adding (2) and (5)
$\log x^z{z}^x=\frac{xyz}{k}\Rightarrow x^z{z}^x=e^{2xyz/k}$
And adding (4) and (6)
$\log y^z{z}^y=\frac{xyz}{k}\Rightarrow y^z{z}^y=e^{2xyz/k}$
Therefore,
$x^y{y}^x+y^z{z}^y-{2x}^z{z}^x=e^{2xyz/k}+e^{2xyz/k}-2e^{2xyz/k}=0$