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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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