Question

Find $x$, if ${\left(\frac{2}{3}\right)}^x{\left(\frac{3}{2}\right)}^{2x}=\frac{81}{16}$.

Answer 1

${\left(\frac{2}{3}\right)}^x{\left(\frac{3}{2}\right)}^{2x}={\left(\frac{3}{2}\right)}^x$

${\left(\frac{3}{2}\right)}^x=\frac{3^x}{2^x}=\frac{81}{16}$

Hence, $3^x=81k$ and $2^x=16k.$

2 and 3 are coprime numbers. Therefore, $k=1.$

We have, $3^x=81$ and $2^x=16.$

Thus the solution is $x=4.$