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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