Why is AdaBoost used to minimize the exponential loss function? I am probably missing something here. 
My understanding is that adaboost attempts to minimize the exponential loss function
$$l(x)=\Sigma_i e^{y*f(x_i)}$$
However, isnt the function that minimizes the exponential loss function already known?
Isnt it $f(x)=0.5\frac{logP(Y=1|x)}{logP(Y=-1|x)}$?
Why canw use f(x) instead of the result produced from adaboost?
 A: I think your understanding is fine but it has not crystallised how the estimates $\text{Pr}(Y=\{-1,+1\}|x)$ came to be and their relation to the expected exponential loss $E\{l(x)\}\}$. We use $l(x) = \exp\{-Yf(x)\}$ exactly because it is minimised at $\frac{1}{2} \log\frac{\text{Pr}(Y=+1|x)}{\text{Pr}(Y=-1|x)}$.
We know by definition that:
$E\{ \exp\{-Yf(x)\}\}$ $=$ $\text{Pr}(Y=+1|x)\exp\{-f(x)\}$ $+$ $\text{Pr}(Y=-1|x)\exp\{f(x)\}$.
Taking the derivative of this is simply:
$-\text{Pr}(Y=+1|x)\exp\{-f(x)\}$ $+$ $\text{Pr}(Y=-1|x)\exp\{f(x)\}$
so setting this to zero directly leads to $\text{Pr}(Y=+1|x) = \frac{\exp\{f(x)\}}{ \exp\{-f(x)\} + \exp\{f(x)\}}= \frac{1}{1+ \exp\{-2f(x)\}}$,
i.e. the one-half of the log-odds of $\text{Pr}(Y=\{-1\}|x)$, which (as Efron & Hastie (2016), (pp. 343) put it): "a perfectly reasonable (and symmetric) model for a probability". i.e. we use $l(x)$ because through $f(x)$ defines to re-express our classification task in probabilistic terms; $f(x)$ independently is not interpretable as sample-wide loss function.
In case you have not come across it yet, Friedman et al. (2000) Additive Logistic Regression: A statistical view of boosting is non-trivial good reference on the matter. We can focus on the Sect. 4 "AdaBoost: an additive logistic regression model" (and Sect. 1 "Introduction"), it should be enough to clear things fully on this matter. A more conversational approach can be found in: Efron & Hastie (2016) Computer Age Statistical Inference, Chapt. 17 "Random Forests and Boosting".
A: Given some observations, how would you calculate the true P(y=1|x) underlying your observed data? If you have a better way of calculating it, yes, you should plug this into $f(x) =  \frac{P(Y=1|x)}{2P(Y=-1|x)}$. If not, AdaBoost is providing a framework to estimate $f(x)$.
Take square loss $L(x,Y) = (Y-f(x))^2$. The function that minimises its expected value is $f(x)=E[Y|x]$. However how would you calculate it? If you assume that $Y$ is in a linear relationship to $x$ then you have your classical linear regression model. This is a framework to estimate $E[Y|x]$. Similarly, AdaBoost tries to estimate $f(x)$ above, given your observations.
