If $ 2^{x}=3^{y}=12^{z} $, show that $ \frac{1}{z}=\frac{1}{y}+\frac{2}{x} $.

Given:

\( 2^{x}=3^{y}=12^{z} \)

To do: 

We have to show that \( \frac{1}{z}=\frac{1}{y}+\frac{2}{x} \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

Let $2^{x}=3^{y}=12^{z}=k$

This implies,

$2=k^{\frac{1}{x}}, 3=k^{\frac{1}{y}}, 12=k^{\frac{1}{z}}$

$\Rightarrow 2^{2} \times 3=(k^{\frac{1}{x}})^2 \times k^{\frac{1}{y}}$

$\Rightarrow 12=(k^{\frac{1}{x}})^{2} \times (k^{\frac{1}{y}})$

$\Rightarrow k^{\frac{1}{z}}=(k^{\frac{1}{x}})^{2} \times (k^{\frac{1}{y}})$

$\Rightarrow k^{\frac{2}{x}} \times k^{\frac{1}{y}}=k^{\frac{1}{z}}$

$\Rightarrow k^{\frac{2}{x}+\frac{1}{y}}=k^{\frac{1}{z}}$

Comparing both sides, we get,
$\frac{2}{x}+\frac{1}{y}=\frac{1}{z}$

Hence proved.

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Updated on: 10-Oct-2022

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