Solve
$x^{ln x} = e^{(lnx)^{3}}$
I'm looking at the mark scheme but I don't understand what they've done. I'd appreciate it if someone could explain every step.
MS: taking ln of both sides or writing $x=e^{lnx}$
$(lnx)^2=(lnx)^3$
$(lnx)^2(lnx-1)=0$
$x=1, x=e$
Taking logs:
$$\ln{(x^{\ln{x}})} = \ln{(e^{(\ln{x})^{3}})}$$
Remember that $\ln{(e^{a})} = a$:
$$\ln{(x^{\ln{x}})} = (\ln{x})^{3}$$
Recall that $\ln{x^{a}} = a\ln{x}$, so $\ln{(x^{\ln{x}})} = (\ln{x})(\ln{x}) = (\ln{x})^{2}$
Hence $(\ln{x})^{2} = (\ln{x})^{3} \iff (\ln{x})^{3} - (\ln{x})^{2} = 0 \iff (\ln{x})^{2}(\ln{x} - 1)=0 \iff \ln{x} = 0$ or $\ln{x} = 1$
So $x = e^{0} = 1$ and $x = e^{1} = e$.
$$x^{\ln x} = e^{(\ln x)^{3}}$$ $$\ln(x^{\ln x}) = \ln(e^{(\ln x)^{3}})$$ $$(\ln x)^2 =(\ln x)^3$$ $$(\ln x)^3-(\ln x)^2=0$$ $$(\ln x)^2(\ln x-1)=0$$ solutions are $$\ln x=1\iff x=e$$ and $$\ln x=0\iff x=1$$
