• September 16th 2012, 05:44 AM
sharmala
hi everyone...can anyone please solve for me the equation x^5 e^-3Inx - 4x=21....thank u
• September 16th 2012, 05:48 AM
Prove It
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*}