Why is it possible that $f(x)$ has real outputs, but $\exp(\ln(f(x))$ has complex outputs?

Question

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 $\mathbb{R}$.

Answer 1

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.

Answer 2

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.