A) \[{{2}^{-3x}}\]
B) \[{{2}^{8x}}\]
C) \[{{2}^{3x}}\]
D) \[{{2}^{11x}}\]
Correct Answer: A
Solution :
\[\frac{6{{4}^{2x}}\div {{16}^{x}}}{{{128}^{x}}\times {{4}^{2x}}}=\frac{{{4}^{4x}}}{{{2}^{11x}}}=\frac{{{2}^{8x}}}{{{2}^{11x}}}={{2}^{-3x}}\]You need to login to perform this action.
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