Question
If $\text{a}^\text{x}=\text{b}^\text{y}=\text{c}^\text{z}$ and $\text{b}^2=\text{ac},$ then show that $\text{y}=\frac{2\text{zx}}{\text{z}+\text{x}}$
✓
Answer
Let $\text{a}^\text{x}=\text{b}^\text{y}=\text{c}^{\text{z}}=\text{k}$
Then, $\text{a}=\text{k}^{\frac{1}{\text{x}}},\ \text{b}=\text{k}^{\frac{1}{\text{y}}}$ and $\text{c}=\text{k}^{\frac{1}{\text{z}}}$
Now,
$\text{b}=\text{ac}$
$\Rightarrow\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{x}}\times\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^\frac{2}{\text{y}}=\text{k}^{\frac{1}{\text{x}}+\frac{1}{\text{z}}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{1}{\text{x}}+\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{\text{z}+\text{x}}{\text{zx}}$
$\Rightarrow\text{y}=\frac{2\text{zx}}{\text{z}+\text{x}}$
Then, $\text{a}=\text{k}^{\frac{1}{\text{x}}},\ \text{b}=\text{k}^{\frac{1}{\text{y}}}$ and $\text{c}=\text{k}^{\frac{1}{\text{z}}}$
Now,
$\text{b}=\text{ac}$
$\Rightarrow\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{x}}\times\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^\frac{2}{\text{y}}=\text{k}^{\frac{1}{\text{x}}+\frac{1}{\text{z}}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{1}{\text{x}}+\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{\text{z}+\text{x}}{\text{zx}}$
$\Rightarrow\text{y}=\frac{2\text{zx}}{\text{z}+\text{x}}$
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