Question

If $\frac{x\left(y+z-x\right)}{\log x}=\frac{y\left(z+x-z\right)}{\log y}=\frac{z\left(x+y-z\right)}{\log z}=$ and $a.x^y.y^x=b.y^z.z^y=c.z^x.x^z$ then $\frac{a+b}{c}$ equals

Answer 1

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

Or, $logx=\frac{x(y+z-x)}{k}$, 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-

$\left(x^y\right)\left(y^x\right)=\left(y^z\right)\left(z^y\right)=\left(x^z\right)\left(z^x\right)$..........(7)

Also, $a{x}^y{y}^x=b{y}^z{z}^y=c{x}^z{z}^x$..........(8)

Comparing $7$ and $8$, we get:

$a=1,b=1,c=1$

Hence, $\frac{a+b}{c}=\frac{2}{1}=2$