Question

$(c^{-2}{)}^{2/3}\times c^{3/2}÷c^{-3}$ is equal to $c^{\frac{m}{n}}$ . If $m$ snd $n$ are co prime fiind the value of $m-n$.

Answer 1

We have,

${\left(c^{-2}\right)}^{2/3}\times c^{3/2}÷c^{-3}=c^{\frac{m}{n}}$

$c^{-4/3}\times c^{3/2}\times c^3=c^{\frac{m}{n}}$

$c^{-4/3}\times c^{9/2}=c^{\frac{m}{n}}$

$c^{(27-8)/6}=c^{\frac{m}{n}}$

$c^{\frac{19}{6}}=c^{\frac{m}{n}}$

$\therefore m=19,n=6$

So,

$m-n=19-6=13$

Hence, this is the answer.