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...