Question

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

Answer 1

Simplifying the expression
Let, $a^x=b^y=c^z=k$
$a^z=k,b^y=k$ and $c^z=k$
$a=k^{\frac{1}{x}},b=k^{\frac{1}{y}}$ and $c=k^{\frac{1}{z}}$
Now,
$b^2=ac$
Substitute the value of $a,b$ and $c$ we get
${\left(\frac{1}{ky}\right)}^2=k^{\frac{1}{z}}\times k^{\frac{1}{z}}$
$k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}$ $[\therefore a^m\times a^n=a^{m+n}]$
$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$\Rightarrow \frac{2}{y}=\frac{z+x}{xz}$
Reversing both sides, we get
$\frac{y}{2}=\frac{xz}{z+x}$
$\therefore y=\frac{2xz}{x+z}$
Hence, proved.