- Home
- Standard 11
- Mathematics
Basic of Logarithms
easy
If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
A
$0$
B
$1$
C
$n$
D
$2n$
Solution
(d) ${2^{m(n + 1) + 2n + n}} = {2^{(m + 1)n + 2m}}$
$ \Rightarrow $$mn + m + 3n = mn + 2m + n$$ \Rightarrow $$m = 2n$.
Standard 11
Mathematics