MCQ
If $a^x=b^y=c^2$ and $a^3=b^2 c$, then $\frac{3}{x}-\frac{2}{y}=$
- A$\frac{x}{y}$
- B$\frac{y}{x}$
- Cxyz
- ✓$\frac{1}{z}$
✓
Answer
Correct option: D.
$\frac{1}{z}$
(d)
Let $a^x=b^y=c^z=k$. Then, $a=k^{1 / x}, b=k^{1 / y}$ and $c=k^{1 / z}$.
$\therefore \quad a^3=b^2 c \Rightarrow\left(k^{1 / 2}\right)^3=\left(k^{1 / y}\right)^2\left(k^{1 / z}\right) \Rightarrow k^{\frac{3}{x}}=k^{\frac{2}{y}+\frac{1}{z}} \Rightarrow \frac{3}{x}=\frac{2}{y}+\frac{1}{z} \Rightarrow \frac{3}{x}-\frac{2}{y}=\frac{1}{z}$
Let $a^x=b^y=c^z=k$. Then, $a=k^{1 / x}, b=k^{1 / y}$ and $c=k^{1 / z}$.
$\therefore \quad a^3=b^2 c \Rightarrow\left(k^{1 / 2}\right)^3=\left(k^{1 / y}\right)^2\left(k^{1 / z}\right) \Rightarrow k^{\frac{3}{x}}=k^{\frac{2}{y}+\frac{1}{z}} \Rightarrow \frac{3}{x}=\frac{2}{y}+\frac{1}{z} \Rightarrow \frac{3}{x}-\frac{2}{y}=\frac{1}{z}$
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