If $\displaystyle \frac{\left( \frac{2x^2}{3a} \right)^{n-1}} {\left( \frac{3x}{a} \right)^{n+1}} = \left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ and substituted the same to find the value of $n$, to no avail
The right values for $a$ is $\pm(3^6 2^{-11/2})$, and $n$ is equal to $6$
On
Expand the LHS as
$$
\frac{{\left( {\frac{{2x^{\,2} }}
{{3a}}} \right)^{n - 1} }}
{{\left( {\frac{{3x}}
{a}} \right)^{n + 1} }} = \frac{{2^{n - 1} x^{2\left( {n - 1} \right)} }}
{{3^{n - 1} a^{n - 1} }}\;\frac{{a^{n + 1} }}
{{3^{n + 1} x^{n + 1} }} = \frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\;
$$
The equating it to the RHS, you shall impose that the exponent
of $x$ be equal on both sides, as well as the multiplying coefficients.
Therefore:
$$ \begin{gathered} \frac{{2^{n - 1} a^2 x^{n - 3} }} {{3^{2n} }}\; = \frac{{x^3 }} {{4^3 }}\quad \Rightarrow \quad \left\{ \begin{gathered} x^{n - 3} = x^3 \hfill \\ \frac{{2^{n - 1} a^2 }} {{3^{2n} }} = \frac{1} {{4^3 }} \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} n = 6 \hfill \\ \frac{{2^5 a^2 }} {{3^{12} }} = \frac{1} {{4^3 }} \hfill \\ \end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered} n = 6 \hfill \\ a^2 = \frac{{3^{12} }} {{2^5 4^3 }} = \frac{{3^{12} }} {{2^{11} }} \hfill \\ \end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered} n = 6 \hfill \\ a = \pm \sqrt 2 \frac{{3^6 }} {{2^6 }} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$
\begin{align*} \frac{\left( \frac{2x^2}{3a} \right)^{n-1}} {\left( \frac{3x}{a} \right)^{n+1}} &= \left( \frac{x}{4} \right)^3 \\ \frac{2^{n-1}}{3^{2n}a^{2n}} x^{n-3} &= \frac{x^3}{2^6} \end{align*}
Comparing $x$ terms,
$$n-3=3 \implies n=6$$
Comparing coefficient of $x^3$,
$$\frac{2^5 a^2}{3^{12}}=\frac{1}{2^6} \implies a^{2}=\frac{3^{12}}{2^{11}} \implies a=\pm \frac{3^6}{2^{11/2}}$$