TITLE: Silly question on radical
QUESTION [7 upvotes]: I'm dealing with another silly problem, now with radicals, I've the equation:
$(\frac{1}{9})^x = \sqrt{27}$
Working on it:
$9^{-x} = 27^{\frac{1}{2}}$
$9^{-x} = 3^{3\times\frac{1}{2}}$
$9^{-x} = 3^{\frac{3}{2}}$
$3^{-3x} = 3^{\frac{3}{2}}$
with exponents:
$-2x = \frac{3}{2}$
$x = \frac{3}{2} \cdot (-\frac{1}{2})$
$x = -\frac{3}{4}$
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 ?

REPLY [2 votes]: 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.
$(\frac19)^{-\frac34}=[(\frac19)^{-1}]^\frac34=9^\frac34=(9^\frac12)^\frac32=3^\frac32=(3^3)^\frac12=\sqrt{27}$