Question

If $a^x=b^y=c^z=d^w,$ then ${\log }_a\left(bcd\right)-x(\frac{1}{y}+\frac{1}{z}+\frac{1}{w})$ is equal to

Answer 1

$a^x=b^y=c^z=d^w\Rightarrow b=a^{\frac{x}{y}};c=a^{\frac{x}{z}};d=a^{\frac{x}{w}}$

Now ${\log }_a\left(bcd\right)={\log }_a\left(\left(a^{\frac{x}{y}}\right)\left(a^{\frac{x}{z}}\right)\left(a^{\frac{x}{w}}\right)\right)={\log }_a\left(a^{(\frac{x}{y}+\frac{x}{z}+\frac{x}{w})}\right)$
$=(\frac{x}{y}+\frac{x}{z}+\frac{x}{w}){\log }_{a}a=x(\frac{1}{y}+\frac{1}{z}+\frac{1}{w})$

Hence ${\log }_a\left(bcd\right)-x(\frac{1}{y}+\frac{1}{z}+\frac{1}{w})=0$