Find $x$ where $f(e^{(x+1)})=x-\ln(x)$ approaches one.

Question

Given that $$ x \in [1,\infty) \quad f(e^{(x+1)})=x-\ln(x) $$

and

$$ \lim_{x \to a} f(x)=1 $$

Find $a$.

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I got to the point: $$ \ln(a)-\ln(\ln(a)-1)=2 $$

But from there on I could not get to $a=e^2$ which is the answer. (I can see that $a=e^2$ is the answer, I just cannot show it.)

Answer 1

Once you have that $$\frac{a}{e^2}=\ln{\left(\frac{a}e\right)}$$ We can proceed to get $$\exp{\left(\frac{a}{e^2}\right)}=\frac{a}e$$ $$1=\frac{a}e\exp{\left(-\frac{a}{e^2}\right)}$$ $$-\exp{(-1)}=-\frac{a}{e^2}\exp{\left(-\frac{a}{e^2}\right)}$$ Now with the equation in this form we can see that $$-\frac{a}{e^2}=-1$$ $$a=e^2$$ In fact there are infinite complex solutions given by $$a=-e^2W_k\left(-\frac1e\right)$$ where $W_k(z)$ is the $k$th branch of the Lambert-W function.

Answer 2

Substituting $$t=e^{x+1}$$ we get $$x=\ln(t)-1$$ so we get $$f(t)=\ln(t)-1-\ln(\ln(t)-1)$$