Question

If $a^x=b^y=c^z$ and $b^2=ac$, then $\frac{1}{x}+\frac{1}{z}$ is equal to

• A. y
• B. $\frac{1}{y}$
• C. $\frac{2}{y}$
• D. $y^2$

Answer 1

Answer: (C)

$\frac{2}{y}$

Let $a^x=b^y=c^z=k$, so that $a=k^{\frac{1}{x}}$, $b=k^{\frac{1}{y}}$, $c=k^{\frac{1}{z}}$. Also, $b^2=ac$. Therefore, ${\left(k^{\frac{1}{y}}\right)}^2=k^{\frac{1}{x}}\cdot k^{\frac{1}{z}}$, i.e., $k^{\frac{2}{y}}=k^{(\frac{1}{x}+\frac{1}{z})}$ or $\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$