A) 1
B) 0
C) 4
D) 5
Correct Answer: A
Solution :
(a): \[\frac{{{16}^{(2m+1)}}\times {{64}^{5}}}{{{256}^{2}}\times 4}=\frac{{{16}^{(2m+1)}}\times {{\left( {{16}^{3/2}} \right)}^{5}}}{{{(16)}^{4}}\times 4}\]. \[=\frac{{{16}^{(2m+1)}}\times {{64}^{\frac{7}{2}}}}{4}=\frac{{{16}^{(2m+\frac{9}{2})}}}{4};RHS={{(16)}^{6m}}\] Since LHS = RHS; \[\therefore {{16}^{2}}^{m+4}={{16}^{6m}}\Rightarrow 2m+4=6m\Rightarrow m=1\]You need to login to perform this action.
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