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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 $x=e^{lnx}$

$(lnx)^2=(lnx)^3$

$(lnx)^2(lnx-1)=0$

$x=1, x=e$

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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 $x = e^{1} = e$.

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$$x^{\ln x} = e^{(\ln x)^{3}}$$ $$\ln(x^{\ln x}) = \ln(e^{(\ln x)^{3}})$$ $$(\ln x)^2 =(\ln x)^3$$ $$(\ln x)^3-(\ln x)^2=0$$ $$(\ln x)^2(\ln x-1)=0$$ solutions are $$\ln x=1\iff x=e$$ and $$\ln x=0\iff x=1$$

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