If $ 2^{x}=3^{y}=6^{-z} $, show that $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 $.

Given:

 \( 2^{x}=3^{y}=6^{-z} \)

To do: 

We have to show that \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \).

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 $2^{x}=3^{y}=6^{-z}=k$

This implies,

$2=k^{\frac{1}{x}}, 3=k^{\frac{1}{y}}$ and $6=k^{\frac{-1}{z}}$
$\Rightarrow 2 \times 3=k^{\frac{-1}{z}}$

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

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

Comparing both sides, we get,

$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Hence proved. 

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Updated on: 10-Oct-2022

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