If \[{{3}^{2x-y}}={{3}^{x+y}}=\sqrt{27},\] then the value of \[{{3}^{x-y}}\]will be [SSC (CPO) 2015] |
A) \[\frac{1}{\sqrt{27}}\]
B) 3
C) \[\sqrt{3}\]
D) \[\frac{1}{\sqrt{3}}\]
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, \[{{3}^{x-y}}={{3}^{1-\frac{1}{2}}}={{3}^{1/2}}=\sqrt{3}\] |
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