please help me to solve the equation

**sharmala** · September 16th 2012, 04:44 AM

hi everyone...can anyone please solve for me the equation x^5 e^-3Inx - 4x=21....thank u

**Prove It** · September 16th 2012, 04:48 AM

No we will not, that's your job.

A hint: $\displaystyle \begin{align*} e^{-3\ln{x}} = e^{\ln{\left( x^{-3} \right)}} = x^{-3} \end{align*}$ , which means your equation becomes

$\displaystyle \begin{align*} x^5 e^{-3\ln{x}} - 4x &= 21 \\ x^5 \left( x^{-3} \right) - 4x &= 21 \\ x^2 - 4x &= 21 \\ x^2 - 4x - 21 &= 0 \end{align*}$

Surely you can solve this quadratic equation...

**sharmala** · September 16th 2012, 04:57 AM

thank u very much for solution

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