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Exponents Of Real Numbers — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsExponents Of Real Numbers3 Marks
Question
If $2^\text{x}=3^\text{y}=6^{-\text{z}},$ Show that $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=0.$
✓

Answer

Let $2^\text{x}=3^\text{y}=6^{-\text{z}}=\text{k}$ Then, $2=\text{k}^{\frac{1}{\text{x}}},\ \text{3}=\text{k}^{\frac{1}{\text{y}}}$ and $6=\text{k}^{-\frac{1}{\text{z}}}$ Now,$6=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow2\times3=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\text{k}^\frac{1}{\text{y}}=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}+\frac{1}{\text{y}}}=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}=-\frac{1}{\text{z}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=0$

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