If anon's answer seems a little abstract to you, you can argue through inequalities that
for $ \ x < 0 \ , $
$$ \ 0 \ < \ e^x \ < \ 1 \ \Rightarrow \ 1 \ < \ 1 + e^x \ < \ 2 \ $$
$$\Rightarrow \ \ln \ 1 \ = \ 0 \ < \ \ln (1 + e^x) \ < \ \ln 2 \ < 1 \ . $$
So $ \ \ln (1 + e^x) \ $ lies entirely within the bounds of the cosine function (and in fact is always positive), and thus for negative values of $ \ x \ $ , there will be two solutions of $ \ \ln (1 + e^x) \ = \ \cos x \ $ for every "cycle" of the cosine function on $ \ ( -\infty \ , \ 0 ) \ . $
The existence of a positive solution is found through the Intermediate Value Theorem, as anon has indicated. To see that it is unique, consider the derivative of the function $ \ \ln (1 + e^x) - \cos x \ $ for $ \ x > 0 \ $ to show that the function is monotonically increasing.