Exponential Recap III

$\begin{aligned} \sqrt[n]{3^n9^{n-\frac 12}} & = 3^{n+1}\sqrt[n]{9^{2n}} \\ \left(3^n 3^{2n-1}\right)^\frac 1n & = 3^{n+1} \left(3^{4n}\right)^\frac 1n & \small \color{#3D99F6} \text{Raise both sides to the power of }n \\ 3^n 3^{2n-1} & = 3^{n^2+n} 3^{4n} \\ 3^{\color{#3D99F6}3n-1} & = 3^{\color{#3D99F6}n^2+5n} & \small \color{#3D99F6} \text{Equating the powers of both sides.} \\ \color{#3D99F6}3n-1 & = \color{#3D99F6}n^2+5n \\ n^2 + 2n + 1 & = 0 \\ (n+1)^2 & = 0 \\ \implies n & = \boxed{-1} \end{aligned}$

$\sqrt[n]{3^n\cdot 9^{n-\frac{1}{2}}}=3^{n+1}\cdot \sqrt[n]{9^{2n}}$

$\left(3^n\cdot 9^{n-\frac{1}{2}}\right)^{\frac{1}{n}}=3^{n+1}\cdot \left(9^{2n}\right)^{\frac{1}{n}}$

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

$\left(3^n\cdot 3^{2n-1}\right)^{\frac{1}{n}}=3^{n+1}\cdot \left(3^{4n}\right)^{\frac{1}{n}}$

$\left(3^{n+2n-1}\right)^{\frac{1}{n}}=3^{n+1+\left(\frac{1}{n}\cdot 4n\right)}$

$3^{\frac{3n-1}{n}}=3^{n+5}$

$3-\frac{1}{n}=n+5$

$3n-1=n^2+5n$

$n^2+5n-\left(3n-1\right)=0$

$n^2+2n+1=0$

$\left(n+1\right)^2=0$

$∴ n=-1$