Question

If $a^x=bc,b^y=ca,c^z=ab$, then the value of $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}$ is

Answer 1

Given that $a^x=bc$
$\Rightarrow x\log a=\log bc$, taking log on both sides
$\Rightarrow \frac{x}{1+x}=\frac{\log bc}{\log a+\log bc}=\frac{\log bc}{\log abc}[\because \log x+\log y=\log (xy\left)\right]$
Similarly, we can get $\frac{y}{1+y}=\frac{\log ac}{\log abc}$ and $\frac{z}{1+z}=\frac{\log ab}{\log abc}$
Now, $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}=\frac{\log bc+\log ac+\log ab}{\log abc}=\frac{\log a^2{b}^2{c}^2}{\log abc}=2[\because \log a^m=m\log a]$
Ans: 2