A) 0
B) \[\frac{3}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{8}\]
Correct Answer: B
Solution :
\[{{2}^{x+y}}=\sqrt{8},\]\[{{2}^{x-y}}=\sqrt{8}\] \[{{2}^{x+y}}=2\sqrt{2},\]\[{{2}^{x-y}}=2\sqrt{2}\] \[{{2}^{x+y}}={{2}^{3/2}},\]\[{{2}^{x-y}}={{2}^{3/2}}\] On comparing, So, \[x+y=\frac{3}{2},x-y=\frac{3}{2}\] On adding, we get \[2x=3\] \[\therefore \] \[x=\frac{3}{2}\]and y = 0
You need to login to perform this action.
You will be redirected in
3 sec
