I have to prove that $e^x= 2x+a$ for $a>2-2\log(2)$ there are 2 distinct solution. I've thought using Bolzano's theorem but I can't understand where to start, thanks in advance.
Edit: the solution to find are, at least, 2.
Edit: I've tried but I can't figure out how to deal with the parameter, $x = \log(2x + a)$ or alternate form like $x - \ln(e^x - a) - \ln 2 = 0$ but I can't continue.
Let
$$f(x) = e^x - 2x \tag{1}\label{eq1A}$$
First, to find any local minimum or maximum points, check the first derivative being $0$, i.e.,
$$f'(x) = e^x - 2 = 0 \implies x = \ln(2) \tag{2}\label{eq2A}$$
The second derivative is $f''(x) = e^x \gt 0$, so $x = \ln(2)$ is a local minimum.
Note as $x \to -\infty$ that $e^x \to 0$ and $-2x \to \infty$, so $\lim_{x \to -\infty}f(x) = \infty$. Also, $e^x$ grows faster than any polynomial, including $2x$, so we have $\lim_{x \to \infty}f(x) = \infty$.
Thus, the global minimum value for $f(x)$ occurs at $x = \ln(2)$, giving
$$f(\ln(2)) = e^{\ln(2)} - 2\ln(2) = 2 - 2\ln(2) \tag{3}\label{eq3A}$$
This shows that, for all $x$, we have
$$e^x - 2x \ge 2 - 2\ln(2) \iff e^x \ge 2x + (2 - 2\ln(2)) \tag{4}\label{eq4A}$$
Thus, with
$$e^x = 2x + a \tag{5}\label{eq5A}$$
for $a = 2 - 2\ln(2)$, there's only $x = \ln(2)$ which satisfies \eqref{eq5A} while, since $f(x)$ is continuous and $\lim_{x \to \pm\infty}f(x) = \infty$, by the intermediate value theorem, for any $a \gt 2 - 2\ln(2)$, there's at least two values (with it actually being exactly $2$ since $f(x)$ is strictly decreasing for $x \lt \ln(2)$ and strictly increasing for $x \gt \ln(2)$) of $x$ which satisfies \eqref{eq5A}.