So I was again looking through the homepage of Youtube when I came across this video by SyberMath which proposed the question$$\text{Solve: }x^{\ln(x)}=ex$$which I thought that I might be able to solve. Here is my attempt at solving the aforementioned equation:$$x^{\ln(x)}=ex$$
$$\ln\left(x^{\ln(x)}\right)=\ln(e\cdot x)$$
$$\ln(x)\ln(x)=\ln(e)\cdot\ln(x)$$
$$(\ln(x))^2=\ln(x)$$
$$\frac{(\ln(x))^2}{\ln(x)}=1$$
$$\ln(x)=1$$
$$e^{\ln(x)}=e^1$$
$$e=x$$
$$\mathbf{\text{My question}}$$
Is the solution that I have arrived at correct, or what could I do to attain the correct solution/attain it more easily?
Not correct. You failed here: $$\ln(e\cdot x)\neq\ln e\cdot \ln x$$ Here is the solution: $$x^{\ln(x)}=ex$$ $$\ln\left(x^{\ln(x)}\right)=\ln(e\cdot x)$$ $$\ln(x)\ln(x)=\ln(e)+\ln(x)$$ $$\ln^2(x)=1+\ln x$$ $$\ln^2(x)-\ln(x)-1=0$$ now take $\ln x=y$: $$y^2-y-1=0$$ which is clearly solved by $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$. So now you have $$\ln(x)=\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$$ Exponentiate and you get: $$x=e^\phi,\ e^{-\frac{1}{\phi}}$$