# Thread: solving for x (negative exponent)

1. ## solving for x (negative exponent)

Hey!

I need some help with solving this equation for x...

(1+(6^2*x^-2))^-1*(-6x^-2)-(1+(4^2*x^-2))^-1*(-4x^-2) = 0

I have problem with (1+(6^2*x^-2))^-1 and (1+(4^2*x^-2))^-1

thanks

2. $(1+(6^2x^{-2}))^{-1}(-6x^{-2})-(1+(4^2x^{-2}))^{-1}(-4x^{-2}) = 0$

$x^{-a} = \frac 1{x^a}$

$\frac 1{(1+(6^2x^{-2}))} * \frac{-6}{x^{2}}-\frac 1{(1+(4^2x^{-2}))} * \frac{-4}{x^{2}} = 0$

$\frac {-6}{x^{-2}(1+(6^2x^{-2}))} - \frac{-4}{x^{-2}(1 + (4^2x^{-2}))} = 0$

3. Originally Posted by derfleurer
$(1+(6^2x^{-2}))^{-1}(-6x^{-2})-(1+(4^2x^{-2}))^{-1}(-4x^{-2}) = 0$

$x^{-a} = \frac 1{x^a}$

$\frac 1{(1+(6^2x^{-2}))} * \frac{-6}{x^{2}}-\frac 1{(1+(4^2x^{-2}))} * \frac{-4}{x^{2}} = 0$
The two $x^2$ should be $x^{-2}$ shouldn't they?

$\frac {-6}{x^{-2}(1+(6^2x^{-2}))} - \frac{-4}{x^{-2}(1 + (4^2x^{-2}))} = 0$
Okay now the they have become $x^{-2}$.

4. Haha, ok, so I gutted that last line. =p

$\frac {-6}{x^2(1+(6^2x^{-2}))} - \frac{-4}{x^2(1 + (4^2x^{-2}))} = 0$

5. wow, quick responses