I need one clarification for this: $5^{(x^{2})} =2^{x} \cdot e^{(2x)}$

Question

I solved this equation by giving it these 2 solutions Now the system says I am partially correct and gives me this solution instead. Can someone please explain why I am not fully correct on my answer?

Answer 1

I assume you want to solve $5^{(x^2)}-2^xe^{2x}=0$, that is \begin{align*} 5^{(x^2)}-2^xe^{2x}&=0\\ 5^{(x^2)}&=2^xe^{2x}\\ \log\left(5^{(x^2)}\right)&=\log\left(2^xe^{2x}\right)\\ x^2\log\left(5\right)&=\log\left(2^x\right)+\log\left(e^{2x}\right)\\ x^2\log\left(5\right)&=x\log(2)+2x\log(e)\\ x^2\log\left(5\right)&=x\log(2)+2x\\ x^2\log\left(5\right)-x\log(2)-2x&=0\\ x\left(x\log\left(5\right)-\log(2)-2\right)&=0\\ \end{align*} so $x=0$ or $x=\dfrac{\log(2)+2}{\log(5)}$.

WolframAlpha agrees with me, so it seems both your original answer and the proposed one are wrong.