A) n+8 |
B) 2n+10 |
C) 3n+2 |
D) 3n+10 |
D) 3n+10 |
Given that, \( \Large 16 \times 8^{n+2} = 2^{m} \)
=> \( \Large \left(2\right)^{4} \times 2^{3 \left(n+2\right) } = 2^{m} \)
=> \( \Large \left(2\right)^{ \left(4+3n+6\right) } = 2^{m} \)
=> \( \Large 2^{ \left(3n+10\right) }= 2^{m} \)
On comparing, we get
\( \Large 3n + 10 = m \)
=> m = 3n + 10