If \[{{2}^{2x-y}}=16\]and \[{{2}^{x+y}}=32,\] then the value of \[xy\] is |
A) 2
B) 6
C) 4
D) 8
Correct Answer: B
Solution :
\[{{2}^{2x-y}}=16={{2}^{4}}\] |
\[\Rightarrow \] \[2x-y=4\] |
\[\Rightarrow \]\[{{2}^{x+y}}=32={{2}^{5}}\] |
\[\Rightarrow \] \[x+y=5\] |
Solving for \[x\]and \[y,\]\[3x=9\]\[\Rightarrow \]\[x=3\]and \[y=2\] |
\[\therefore \] \[xy=6\] |
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