Schwarzschild solution - why use exponential functions? Often in the literature, when deriving the famous Schwarzschild solution  the metric can be cast in two equivalent ways:
$$(ds)^2 = e^{2\alpha(r,t)} dt^2 - e^{2\beta(r,t)}dr^2 - r^2 d\Omega^2. $$
$$ (ds)^2 = A(r,t)dt^2 - B(r,t)dr^2 - r^2 d\Omega^2. $$
However, I don't necessarily understand why we can write it in the exponential form? There is obviously a very simple answer. Is it something to do with the unknown function never being identically zero... or at least something along these lines?
UPDATE (post comment(s)) 
In response to the comment by ArthurSuvorov. The exponential function is never negative for real input and will keep the manifold Lorentzian. This makes sense. However, what about if we write it in the second form? There isn't conditions on these unknown functions initially? Or is there?
 A: The exponential function is never negative for real input, and one wants to keep the signature of the metric the same [(+,-,-,-) in your case] for all values of $t$ and $r$. Writing the coefficients as exponentials enforces this without the need to worry about domains and ranges so much.
A: For the Schwarzschild solution it is convenient to do so as @Arthur Suvorov pointed out. But in general the second non-exponential ansatz can also be used. Asymptotic flatness 
$$\lim_{r\rightarrow\infty}A(r,t)=1$$
$$\lim_{r\rightarrow\infty}B(r,t)=1$$
poses two boundary conditions for $A$ and $B$ (or $\alpha$ and $\beta$). The field equations or in general the static nature of the problem implies that $A$ and $B$ are functions of $r$ only. One can find the Schwarzschild solution using $A$ and $B$ but $\alpha$ and $\beta$ are nicer to work with.
In fact it is not uncommon to use such an exponential ansatz since they often make computations a lot simpler. 
Inside a spherical symmetric body the exponential form of the $g_{tt}$ becomes very important: Its radial derivative is directly related to the fluids $\log$-enthalpy gradient:
$$\frac{P'(r)}{\rho(r)+P(r)}\equiv h'(r)=-\frac{1}{2}\frac{\partial \ln(g_{tt}(r))}{\partial r}.$$
This is the general relativistic extension of the classical Bernoulli theorem (in the special case of a static, spherical symmetric ideal fluid interior). In the classical limit $\frac{1}{2}\frac{\partial \ln(g_{tt}(r))}{\partial r}$ becomes the classical Newtonian gravitational potential. So in some cases it makes more sense to use the exponential forms apart from pure mathematical convenience.
