Question

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

• A. $0$
• B. $abc$
• C. $a+b+c$
• D. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Answer 1

The correct option is A $0$

$Let{a}^x=b^y=c^z=k$ and $abc=1$ [given]

$a^x=k$, $b^y=k$, $c^z=k$
$a=k^{\frac{1}{x}}$, $b=k^{\frac{1}{y}}$, $c=k^{\frac{1}{z}}$
$k^{\frac{1}{x}}\times k^{\frac{1}{y}}\times k^{\frac{1}{z}}=1$
$k^{[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]}=k^0$
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$