Question : If $6^{x}=3^{y}=2^{z}$, then what is the value of $\frac{1}{y}+\frac{1}{z}-\frac{1}{x}$?
Option 1: 1
Option 2: 0
Option 3: 3
Option 4: 6
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Correct Answer: 0
Solution :
Given,
$6^{x}=3^{y}=2^{z}$
Consider, $6^{x}=3^{y}=2^{z}=k$(say), [$k$ is some variable]
⇒ $6^x=k$
⇒ $6=k^{\frac{1}{x}}$ .....(i)
Similarly,
$3=k^{\frac{1}{y}}$ .........(ii)
And $2=k^{\frac{1}{z}}$ ..........(iii)
Multiplying (ii) and (iii), we get,
⇒ $3\times 2=k^{\frac{1}{y}}\times k^{\frac{1}{z}}$
⇒ $6=k^{\frac{1}{y}+\frac{1}{z}}$
Dividing by (i), we get,
⇒ $\frac66=\frac{k^{\frac{1}{y}+\frac{1}{z}}}{k^{\frac{1}{x}}}$
⇒ $1=k^{\frac{1}{y}+\frac{1}{z}-\frac1x}$
⇒ $\frac{1}{y}+\frac{1}{z}-\frac1x=0$
Hence, the correct answer is 0.
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