$9^x=5$, 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.

You don't need to do that at all. $$5\cdot 27^{-x-1}=\dfrac{5}{27^{x+1}}=\dfrac{5}{27}\dfrac{1}{27^x}=\dfrac{5}{27}\dfrac{1}{(9^x)^{1.5}}=\dfrac{5}{27}\dfrac{1}{5\sqrt 5}=\dfrac{1}{27\sqrt 5}$$
