Question

if $2^x=3^y=6^{-z}$ then $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ is equal to

• A. 0
• B. 1
• C. $\frac{3}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A 0

$2^x=3^y=6^{-z}$
Let, $2^x=3^y=6^{-z}=k$
therefore, $x=\frac{\log k}{\log 2}$, $y=\frac{\log k}{\log 3}$, $z=-\frac{\log k}{\log 6}$
Now, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\log 2}{\log k}+\frac{\log 3}{\log k}-\frac{\log 6}{\log k}$
= $\frac{\log 2}{\log k}+\frac{\log 3}{\log k}-\frac{\log 6}{\log k}$
= $\frac{\log 2+\log 3-\log 6}{\log k}$
= $\frac{\log \frac{2\times 3}{6}}{\log k}$
= $\frac{\log 1}{\log k}$
= $0$