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 $\mathbb{R}$.