$9^x=5$, solve for $5\cdot27^{-x-1}$ Question:
Given $9^x=5$
Compute the value of $5\cdot27^{-x-1}$
My attempt:
$9^x=(3^x)^2=5\Rightarrow3^x=\sqrt{5}$
$5\cdot27^{-x-1}=3^{2x}\cdot(3^{3-x}\div3^{3})=3^{2x}\cdot3^{-x}=3^{x}=\sqrt{5}$
My answer is incorrect and I wonder why.

If you want to use your result

$$5.27^{-x-1}=\frac{9^x}{27.27^{x}}=\frac{1}{27}\frac{1}{3^x}=\frac{1}{27 \sqrt{5}}$$

