Why is it possible that $f(x)$ has real outputs, but $\exp(\ln(f(x))$ has complex outputs? This is probably a trivial question, but I can't seem to wrap my head around it.
Say I have a function: $f(x)=x(x-6)^{1/5}$. Is the output of the function just the real numbers?
Desmos tells me yes, but my friend told me that it isn't, because $f(x)=\exp(\ln(x(x-6)^{1/5})=\exp(\ln(x)+\frac15\ln(x-6))$, and because $\ln$ of a negative number will yield a complex output, $f(x)$ should give complex answers as well. 
Plotting $f(x)=\exp(\ln(x)+\frac15\ln(x-6))$, I see that it only gives real values from 6 to infinity.
Or does this mean that $x(x-6)^{1/5}$and $\exp(\ln(x)+\frac15\ln(x-6))$ are actually different functions after all?
I thought about when $(-1)^n$ is imaginary and real. When $n=0.2$, $(-1)^n=-1$ and so my function $f(x)$ should be able to handle negative $x$'s perfectly well.
However, I found that $f(x)=x(x-6)^{1/10}$ outputs real values only from 6 to infinity. This is probably because $(-1)^n$ is imaginary when $n=0.1$. 
This is my question. Why is my friend's logic sound when $f(x)=x(x-6)^{1/10}$ but it breaks down when $f(x)=x(x-6)^{1/5}$? Does it have anything to do with when $(-1)^n$ is real or is there something else about the exponential/logarithmic function that I don't know about?
I feel like this might be in the realm of complex analysis, but I haven't studied that topic yet.
Edit: Let's say the domain of $f$ is $\mathbb{R}$.
 A: In the complex field the logarithm is a multivalued function, let's designate it by $Log$:
$$
Log(z) = \ln (\left| z \right|) + i\,\arg \left( z \right) + i2k\pi 
$$
When you take the principal branch of that function
$$
\log (z) = \ln (\left| z \right|) + i\,\arg \left( z \right)
$$
and apply it to a real $z$, you get
$$
\log (x) = \left\{ {\matrix{
   {\ln (\left| x \right|) + i\,\pi } & {x < 0}  \cr 
   {\ln (\left| x \right|)} & {0 < x}  \cr 
 } } \right.
$$
Understanding by $e^z$ the exponential function $e^{|z|} (\cos (\arg(z))+i \sin(\arg(z))$, then you have for instance
$$
\eqalign{
  &  - 1 = e^{\,\log ( - 1)}  = e^{\,0} \left( {\cos \pi  + i\sin \pi } \right) =  - 1  \cr 
  & \left( { - 1} \right)^{\,1/5}  = e^{\,\,1/5\,\log ( - 1)}  = \cos \pi /5 + i\sin \pi /5 = i^{\,2/5}  \cr} 
$$
That, keeping on the principal branch of the logarithm.
But $ \left( { - 1} \right)^{\,1/5} $ has five different solutions in total, which correspond to the various branches of the multivalued $Log$.
If you take the branch corresponding to $k=2$ you get
$$
\left( { - 1} \right)^{\,1/5}  = e^{\,\,1/5\,\ln 1 + 1/5i5\pi }  = e^{\,\,i\pi }  =  - 1
$$
But you do not have any branch that can provide $\left( { - 1} \right)^{\,1/10} =-1$.
The matter is that if you choose a branch of the logarithm and remain within it, then you can reverse - to a certain extent- 
the exponentiation keeping the results congruent, while if you "jump among the branches" you end with incongruencies.
A: The answer (to the titular question) is simple. Whereas the equation $$e^{\log x}=x$$ is true when $x>0$ (and therefore $\log x$ is well-defined and real), it fails to be true when $x<0.$
In more detail, we provide answers to the body of your question. You say,

Say I have a function: $f(x)=x(x-6)^{1/5}.$ Is the output of the function just the real numbers?

Yes, $f(x)$ always takes a real value whenever $x$ is real, since we can always extract the fifth root of any real number (which is again always real, by definition), and the product of any two real numbers is again a real number.
Then your friend's redefinition of $f(x),$ and the attendant consequences, is explained by my answer above to your titular question.
