Question: If $L \times T^{-2} = L^{n+m} \times T^{-m}$ Solve for m , n?
solving for m:
$-2=-m$ (how???)
$m=2$
solving for n:
$n+m=1$ (how???)
$n+2=1$
$n=1-2=-1$
my question is how can we forget about the two L's and just say that $T^{-2}=T^{-m} $
and do the same again this time with the L and ignore the T and say that $L=L^{n+m}$?
Given that $$ L^1 \times T^{-2} = L^{n + m} \times T^{-m} \tag{1} $$ where $L$ and $T$ stand for length and time, respectively.
Equating the like exponents in (1), we get $$ n + m = 1 \tag{2} $$ and $$ -m = -2 \tag{3} $$
From (3), $m = 2$.
Substituting the value of $m$ into (2), we get $$ n = 1 - 2 = -1 $$
Hence, $$ \boxed{m = 2, \ \ n= -1} $$