$9^x=5$, solve for $5\cdot27^{-x-1}$

Question

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.

Answer 1

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

Answer 2

If you want to use your result

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