Approximately solve the equation. Find the first two terms of the approximation. Approximately solve the equation. Find the first two terms of the approximation.
By $a>>1$ and $a<<1$
$\ln x=e^{-ax}$
$x=e^{e^{-ax}}$
$x_{0}\sim1$
$x_{1}=e^{e^{-a}}$
$\left|e^{e^{-a}}-1\right|<<1$
$x=1+e^{-a}+\frac{e^{-2a}}{2}$
Have I considered one case correctly and how to proceed for the second?
 A: I got a bit different answer.
When we use the perturbation theory to find an approximate solution, we search for the solution in the form $x=x_0+x_1+x_2+...$. We also have to check at every iteration that $x_0\gg x_1\gg x_2 ...$

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*$\ln x=e^{-ax},\,\, a\gg1$.

We are looking for the solution in the form $x=1+\epsilon; \,\,\epsilon\ll1$
$$\ln(1+\epsilon)=\epsilon-\frac{\epsilon^2}{2}+ ... =e^{-a}e^{-a\epsilon}$$
Let's also suppose that $a\epsilon\ll1$ (we will have to check whether it is true). Decomposing the exponent
$$=\epsilon-\frac{\epsilon^2}{2}+ ... =e^{-a}(1-a\epsilon+\frac{a^2\epsilon^2}{2}-...)$$
Taking in turn $\epsilon =e^{-a}+\delta; \,\,\delta\ll e^{-a}$ (we also see that the requirement $a\epsilon\sim ae^{-a}\ll1$ is met for $a$ big enough).
$$e^{-a}+\delta-\frac{(e^{-a}+\delta)^2}{2}+...=e^{-a}\Big(1-a(e^{-a}+\delta)+\frac{a^2(e^{-a}+\delta)^2}{2}-...\Big)$$
In this equation we have to keep all terms $\sim e^{-2a}$
$$\delta-\frac{e^{-2a}}{2}-e^{-a}\delta+ ... =-ae^{-2a}-ae^{-a}\delta + ...$$
$\delta \ll e^{-a}$, so we have to drop in the equation all the terms $\sim e^{-a}\delta$
$$\delta-\frac{e^{-2a}}{2}=-ae^{-2a}\,\,\Rightarrow\,\, \delta=e^{-2a}\Big(\frac{1}{2}-a\Big)\ll e^{-a}$$
Therefore, for $\mathbf{a\gg1}$ the solution looks
$$\boxed{x=1+e^{-a}+e^{-2a}\Big(\frac{1}{2}-a\Big)+o(e^{-2a})}$$
2. $\ln x=e^{-ax},\,\, a\ll1$
We are looking for the solution in the form $x=e+\epsilon; \,\,\epsilon\ll1$
$$\ln(e+\epsilon)=\ln e+\ln\Big(1+\frac{\epsilon}{e}\Big)=1+\frac{\epsilon}{e}-\frac{\epsilon^2}{2e^2}+...=1-a(e+\epsilon)+\frac{a^2}{2}(e+\epsilon)^2+...$$
Choosing in turn $\epsilon=-ae^2+\delta,\, \delta\ll a$, and keeping only the terms $\sim a^2$, we get the equation for $\delta$
$$\frac{\delta}{e}-\frac{a^2e^4}{2e^2}=a^2e^2+\frac{a^2}{2}e^2\,\,\Rightarrow\,\, \delta=2e^3a^2$$
For $\mathbf{a\ll1}$ the solution looks
$$\boxed{x=e-e^2a+2e^3a^2+o(a^2)}$$
