Solving $(\frac{1}{9})^x = \sqrt{27}$ for $x$

Question

I'm dealing with another silly problem, now with radicals, I've the equation:$$(\frac{1}{9})^x = \sqrt{27}.$$

Working on it: \begin{align*} 9^{-x} & = 27^{\frac{1}{2}}\\ 9^{-x} & = 3^{3\times\frac{1}{2}}\\ 9^{-x} & = 3^{\frac{3}{2}}\\ 3^{-3x} & = 3^{\frac{3}{2}} \end{align*} with exponents: \begin{align*} -2x & = \frac{3}{2}\\ x & = \frac{3}{2} \cdot (-\frac{1}{2})\\ x & = -\frac{3}{4} \end{align*}

But it's not compatible, where's the mistake?

Edit

Sorry, it was typo, it's fixed now, but I'm still unable to prove that$$\dfrac{1}{9}^{-\dfrac{3}{4}} = \sqrt{27}.$$

How could I do that ?

Answer 1

To answer the new question, you may want to manipulate the exponent in the same fashion you used to get the answer in the first place.

$(\frac19)^{-\frac34}=[(\frac19)^{-1}]^\frac34=9^\frac34=(9^\frac12)^\frac32=3^\frac32=(3^3)^\frac12=\sqrt{27}$

Answer 2

You made a mistake on one of the last lines (also, in the line before "with exponents" it's $3^{-2x}=3^{\dfrac{3}{2}}$ but I think it's a typo). When you have $x=-\dfrac{1}{2}.\dfrac{3}{2}$ the product should be $x=-\dfrac{3}{4}$.