Question: If $ {3^{2x-y}}={3^{x+y}}=\sqrt{27}, $ then the value of $ {3^{x-y}} $ will be
Options:
A) $ \frac{1}{\sqrt{27}} $
B) 3
C) $ \sqrt{3} $
D) $ \frac{1}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ {3^{2x-y}}={3^{x+y}}=\sqrt{27} $
$ \Rightarrow $ $ {3^{2x-y}}={3^{3/2}} $
Now, $ 2x-y=\frac{3}{2} $ … (i)
and $ {3^{x+y}}={3^{3/2}} $
$ \Rightarrow $ $ x+y=\frac{3}{2} $ … (ii)
On adding Eqs. (i) and (ii), we get
$ 3x=3 $
$ \Rightarrow $ $ x=1 $ and $ y=\frac{1}{2} $
Hence, $ {3^{x-y}}={3^{1-\frac{1}{2}}}={3^{1/2}}=\sqrt{3} $