Question

Find the value of $n$ :
${\left(\frac{3}{2}\right)}^6\times {\left(\frac{3}{2}\right)}^{2n+1}\times {\left(\frac{3}{2}\right)}^{-3}=1$

Answer 1

Now,

${\left(\frac{3}{2}\right)}^6\times {\left(\frac{3}{2}\right)}^{2n+1}\times {\left(\frac{3}{2}\right)}^{-3}=1$

or, ${\left(\frac{3}{2}\right)}^{2n+1+6-3}={\left(\frac{3}{2}\right)}^0$

or, $2n+4=0$

or, $n=-2$.