Question
If $3^\text{x}=5^\text{y}=(75^\text{z}),$ show that $\text{z}=\frac{\text{xy}}{2\text{x}+\text{y}}.$
✓
Answer
Let $\text{3}^\text{x}=\text{5}^\text{y}=\text{(75)}^{\text{z}}=\text{k}$ Then, $\text{3}=\text{k}^{\frac{1}{\text{x}}},\ \text{5}=\text{k}^{\frac{1}{\text{y}}}$ and $\text{75}=\text{k}^{\frac{1}{\text{z}}}$ Now,$75=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow3\times5^2=\text{K}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\text{k}^{\frac{2}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}+\frac{2}{\text{y}}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{2}{\text{y}}=\frac{1}{\text{z}}$
$\Rightarrow\frac{\text{y}+2\text{x}}{\text{xy}}=\frac{1}{\text{z}}$
$\Rightarrow\text{z}=\frac{\text{xy}}{2\text{x}+\text{y}}$
$\Rightarrow3\times5^2=\text{K}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\text{k}^{\frac{2}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}+\frac{2}{\text{y}}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{2}{\text{y}}=\frac{1}{\text{z}}$
$\Rightarrow\frac{\text{y}+2\text{x}}{\text{xy}}=\frac{1}{\text{z}}$
$\Rightarrow\text{z}=\frac{\text{xy}}{2\text{x}+\text{y}}$
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