Question

$a^x=b^y=c^x$ and $b^2=ac$ prove that $y=\frac{2xz}{x+z}$

Answer 1

Given,

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

$\to a^x=b^y=c^z=k$

$a=k^{\frac{1}{x}}$

$b=k^{\frac{1}{y}}$

$c=k^{\frac{1}{z}}$

$b^2=ac$

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

$k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}$

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

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

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

Hence Proved.