If $ a^{x}=b^{y}=c^{z} $ and $ b^{2}=a c $, then show that $ y=\frac{2 z x}{z+x} $.

Given:

\( a^{x}=b^{y}=c^{z} \) and \( b^{2}=a c \)

To do: 

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

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 $a^{x}=b^{y}=c^{z}=k$

This implies,

$a=k^{\frac{1}{x}}, b=k^{\frac{1}{y}}$ and $c=k^{\frac{-1}{z}}$

$b^{2}=ac$

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

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

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

$\Rightarrow \frac{2}{y}=\frac{z+x}{xz}$

$\Rightarrow y=\frac{2 xz}{z+x}$

Hence proved.  

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Updated on: 10-Oct-2022

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