ABC26GN1987 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: 2005 · Marks: · Difficulty:
If $3^{2 x-y}=3^{x+y}=\sqrt{27}$, the value of $y$ is
(a)$\frac{1}{2}$
(b)$\frac{1}{4}$
(c)$\frac{3}{2}$
(d)$\frac{3}{4}$
Answer
Explanation
$3^{2 x-y}=3^{x+y}=\sqrt{3^{3}}=3^{\frac{3}{2}} \Leftrightarrow 2 x-y=\frac{3}{2}$ and $x+y=\frac{3}{2}$ $$\begin{aligned} & \Leftrightarrow \quad 3 x=\frac{3}{2}+\frac{3}{2}=3 \Leftrightarrow x=1 . \\ & \therefore \quad y=\left(\frac{3}{2}-1\right)=\frac{1}{2} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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