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

No we will not, that's your job.

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

\displaystyle \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...