Find the domain of $ \large e^{\ln\frac{x}{x+8}} $ Find the domain of $ \large  e^{\ln\frac{x}{x+8}} $
My solution:
Set $f(x) =  \large  e^{\ln\frac{x}{x+8}}$, Now, we know $ e^{\ln x} = x$. So,$f(x) = \frac{x}{x+8}$ 
But again for the domain, since the $\ln $ cannot take negative values we have to assume that $\frac{x}{x+8} \gt 0$, which gives $ \frac{x}{x+8} \gt 0 \implies  x<-8 \text{ or }  x\gt 0$ 
but, according to someone I know it's actually,  $\frac{x}{x+8}\neq 0 $.
But that mean we are operating the $e$ function first and then the rest, but why shouldn't we take care of the $\ln $ first? where exactly I am going wrong?
 A: You are doing things OK. When considering the sign of $$y = \frac{x}{{x + 8}}$$ you have to make two systems of inequalities, namely:$$x > 0 \wedge x + 8 > 0$$ and $$x < 0 \wedge x + 8 < 0$$
There is no $\leq$ or $\geq$ since we don't want the $\log$ to blow up at $0$. This two inequalities will let you see $$\mathbb{D} = ( - \infty , - 8)\cup (0,\infty ) $$
That is, either both numerator and denominator are $+$ thus "$+\times+=+$" or both are negative, thus "$-\times-=+$"
A: You are correct; your friend is incorrect.
But, you have one bad phrasing:
$$
\text{"since the }\ln\text{ cannot take negative values we have to assume that }\textstyle{x\over x+8}>0 ".
$$
is not what you want to say. The $\ln$ can take negative values (for instance $\ln({1\over 1+8})$) and that would be ok; you could  still take $e$ raised to that power
($e^{\ln({1\over 1+8})}= {1\over 1+8} $).  
The problem is that the logarithm of a nonpositive number is not defined. So, what you want to say is:


$$
\text{
"since   the $\ln$ of a nonpositive number is undefined
we have to assume that }\textstyle{x\over x+8}>0"$$

And, incidentally, it is not a good idea to start off with simplifying  $e^{\ln({x\over x+8})}={x\over x+8}$.  This "hides the bad points" when you just look at the right hand side; and in fact this equality is valid only then $x$ is in the domain of your  function. 
A: Your argument is (almost) correct.  Since the domain is the reals, in order for the expression $e^{\log\left(\frac{x}{x+8} \right)}$ to be defined, we need for
$$
\frac{x}{x+8} > 0
$$
which occurs when $x \in (-\infty, -8) \cup (0, \infty)$.  Note that $x = 0$ is not allowed since $\ln(0)$ is undefined.
The second answer would be correct one was finding the domain for the complex numbers.  There, $\log(z) = \ln(z) + i Arg(z)$ is defined (although one gets in to the sticky business of dealing with branche cuts) whenever $z \neq 0$.  So the domain over the complex numbers is $\mathbb{C} \setminus \{0,-8\}$.
A: You are right. Even though 
$$e^{\ln \frac{x}{x+8}}=\frac{x}{x+8}$$
everywhere that the left-hand side is defined, that does not mean that the left-hand side is defined when $\frac{x}{x+8}$ is negative.
Imagine that you are a simple scientific calculator. Mine puts an "E" in the display if I try to divide by $0$, or try to take the $\ln$ of a negative number, or ask it to compute $\sin^{-1}$ of a number $>1$.  The calculator then refuses to go on until it is reset.
To find the largest set of reals at which an expression $\mathcal{E}(x)$ defines a function, pretend that you are a calculator, and are evaluating the expression exactly as it is written. If, for the real number $x$, at some stage of the calculation, the calculator would refuse to go on,  then $\mathcal{E}(x)$  does not define a function at $x$. (I am lying a bit. My calculator also gets upset when it is asked to evaluate $2^{500}$.) 
