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 ?


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 ?

  • 5
    $\begingroup$ Looks like you have a typo on the line before "with exponents", but it looks like the only real mistake is in the final multiplication. $\endgroup$ – Mike Mar 27 '12 at 0:40
  • $\begingroup$ (+1) for work shown. @Mike: You should make that an answer! $\endgroup$ – The Chaz 2.0 Mar 27 '12 at 0:42

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.


  • $\begingroup$ Thank you so much, I'm so bad working with rationals, thanks; $\endgroup$ – aajjbb Mar 28 '12 at 2:59

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 $x=-\dfrac{3}{4}$.


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