Question

If $a^x=b^y=c^z$ and $b^2=ac,$ Prove that y=$\frac{2xz}{Z+x}$

Answer 1

Let $a^x=b^y=c^z=k$

$\therefore a=k^{\frac{1}{x}};b=k^{\frac{1}{y}};c=k^{\frac{1}{z}}$

Also, we have $b^2=ac$

$\therefore {\left(k^{\frac{1}{y}}\right)}^2=\left(k^{\frac{1}{x}}\right)\times \left(k^{\frac{1}{z}}\right)$

$\Rightarrow k^{\frac{1}{y}}=k^{\frac{1}{x}+\frac{1}{z}}$

$\Rightarrow k^{\frac{2}{y}}=k^{\frac{z+x}{kz}}$

Comparing the powers we have

$\frac{2}{y}=\frac{z+x}{xz}$

$\Rightarrow y=\frac{2xz}{z+x}$