solving for x (negative exponent)

**JazzCake** · May 18th 2009, 10:08 AM

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

**derfleurer** · May 18th 2009, 10:15 AM

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

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

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

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

**HallsofIvy** · May 18th 2009, 10:36 AM

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 $x^{-2}$ .

**derfleurer** · May 18th 2009, 10:44 AM

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

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

**JazzCake** · May 18th 2009, 11:05 AM

wow, quick responses

Thank you so much guys. your answers helped me a lot.

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