Question
If $2^\text{x}=3^\text{y}=12^\text{z},$ show that $\frac{1}{\text{z}}=\frac{1}{\text{y}}+\frac{2}{\text{x}}.$
✓
Answer
Let $2^\text{x}=3^\text{y}=12^\text{z}=\text{k}$ T
hen, $2=\text{k}^{\frac{1}{\text{x}}},\ \text{3}=\text{k}^{\frac{1}{\text{y}}}$ and
$12=\text{k}^{\frac{1}{\text{z}}}$
Now,$12=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\Big(\text{k}^{\frac{1}{\text{x}}}\Big)^2\times\text{k}^\frac{1}{\text{y}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^\frac{2}{\text{x}}\times\text{k}^\frac{1}{\text{y}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{2}{\text{x}}+\frac{1}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{\text{z}}$
hen, $2=\text{k}^{\frac{1}{\text{x}}},\ \text{3}=\text{k}^{\frac{1}{\text{y}}}$ and
$12=\text{k}^{\frac{1}{\text{z}}}$
Now,$12=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\Big(\text{k}^{\frac{1}{\text{x}}}\Big)^2\times\text{k}^\frac{1}{\text{y}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^\frac{2}{\text{x}}\times\text{k}^\frac{1}{\text{y}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{2}{\text{x}}+\frac{1}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{\text{z}}$
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