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The question is $$ 4^{\log_2(\ln(x))} = \ln(x)-\ln(x)^2+1$$ Upon simplifying we'd get \begin{align} &\phantom{{}\implies {}} 2\ln(x)^2-\ln(x)-1=0 \\ &\implies \ln(x)=1 \text{ or }\ln(x)={-1 \over 2} \\ &\implies x=e\text { or } x=e^{{-1\over2}} \end{align} I understand that upon putting $e^{-1/2}$ into the original expression, it doesn't give a meaningful result; but is there a better way to explain this?

Bernard
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Infinity 207
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    The main gist of it is that the simplification you use on the left-hand side at the very beginning gets rid of a restriction that $\ln(x) > 0.$ So although the original equation being satisfied implies that $x$ is one of the two values we get at the end, it is not the case that $x$ being one of the two values implies that the equation is satisfied unless you reintroduce the restriction $\ln(x) > 0 \Leftrightarrow x > 1.$ – Stephen Donovan Jun 25 '21 at 09:48
  • Terribly sorry. I thought it had something to do with the range of the function. Sleep-deprived me forgot to consider that $ln(x)$ has to be positive for the function to be defined for real values. – Infinity 207 Jun 25 '21 at 09:55
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    When $x$ is greater than $1$, we have $4^{\log_2(\ln x)}=(\ln x)^2$. It turns out that the expression on the RHS is defined for all $x>0$ (not just $x>1$), whereas the LHS is only defined for $x>1$. This similar to how $\frac{x}{x}=1$ for all $x \ne 0$, but the RHS is actually defined for all $x$. This means that it is possible to get extraneous solutions (solutions that solve $2(\ln x)^2-\ln(x)-1=0$, but not the original equation). – Joe Jun 25 '21 at 09:59
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    @Infinity207 Don't worry about it, I do plenty of dumb things on here when I haven't slept. Glad we got it all figured out – Stephen Donovan Jun 25 '21 at 21:07

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It is $$4^{\log_2(\ln(x))}$$ only defined if $$\ln(x)>0$$ and this is the case if $$x>1$$

Dr. Sonnhard Graubner
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