Question:
If $a^{x}=b^{y}=c^{2}$ and $b^{2}=a c$, then show that
$y=\frac{2 z x}{z+x}$
Solution:
Let $a^{x}=b^{y}=c^{z}=k$
$a=k^{1 / x}, b=k^{1 / y}, c=k^{1 / z}$
Now
$b^{2}=a c$
$\left(k^{1 / y}\right)^{2}=k^{1 / x} \times k^{1 / z}$
$k^{2 / y}=k^{1 / x+1 / z}$
$2 / y=1 / x+1 / z$
$2 y=\frac{x+z}{x z}$
$y=\frac{2 x z}{x+z}$
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