# Question #e6807

Oct 3, 2015

I found ii)

#### Explanation:

We can write:
$A = {B}^{n} \cdot {C}^{m}$ as:
$L T = {\left({L}^{2} {T}^{-} 1\right)}^{n} \cdot {\left(L {T}^{2}\right)}^{m}$
$L T = {L}^{2 n} {T}^{- 1 n} \cdot {L}^{m} {T}^{2 m}$ and:
$L T = {L}^{2 n + m} {T}^{- n + 2 m}$
So to have the two sides equal we need that:
$2 n + m = 1$
$- n + 2 m = 1$
from the second: $n = 2 m - 1$
into the first:
$2 \left(2 m - 1\right) + m = 1$
$4 m - 2 + m = 1$
and:
$5 m = 3$
$m = \frac{3}{5}$
and back into $n = 2 m - 1$ you get:
$n = \frac{6}{5} - 1 = \frac{1}{5}$