Question

If $a^x=b^y=c^z$ and $b^2=ac$ then show that $y=\frac{2zx}{z+x}$

Answer 1

Given : $a^x=b^y=c^z$ and $b^2=ac$
To prove : $y=\frac{2zx}{z+x}$
Let $a^x=b^y=c^z=k$
then $a=k^{\frac{1}{x}};b=k^{\frac{1}{y}};c=k^{\frac{1}{z}}$

Since, $b^2=ac$
$\Rightarrow {\left(k^{\frac{1}{y}}\right)}^2=\left(k^{\frac{1}{x}}\right)\times \left(k^{\frac{1}{z}}\right)$

$\Rightarrow \left(k^{\frac{2}{y}}\right)=k^{\frac{1}{x}+\frac{1}{y}}$
$\Rightarrow k^{\frac{2}{y}}=k^{\frac{x+z}{xz}}$
On comparing both the sides , we get

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

$\Rightarrow y=\frac{2zx}{x+z}$
Hence, proved .