I tried to find all solutions for this equation:
$$X^Y-Z^X=X^X$$
My math is not that good, I thought I could do it with logarithms but this did not work. I already have one set of solutions:
$$X=3,\ Y=5,\ Z=6$$
but I what to know if there are more solutions to this equation and how I should tackle such problem
On
What I would do is make $Y$ the subject, i.e.
\begin{align} X^Y-Z^X = X^X &\Rightarrow X^Y = X^X + Z^X \Rightarrow Y\ln (X) = \ln\left(X^X + Z^X\right)\\ &\Rightarrow Y = \frac{\ln\left(X^X + Z^X\right)}{\ln (X)} \end{align}
Now if you plug in values of $X$ and $Z$, it will provide the corresponding $Y$ value that satisfies your equation.
I guess $x,y,z$ are all positive integers. So, we have: $$x^y-x^x=z^x \implies x^x(x^{y-x}-1)=z^x$$ We have $y>x$ and $x \mid z$. Let $a=y-x$ and $b=\frac{z}{x}$. Thus: $$x^a-1=b^x$$ Thus, we must either have:
Case 1: $a,x>1$
By Mihailescu's theorem, we will have $(x,a,b)=(3,2,2)$. This gives us $(x,y,z)=(3,5,6)$.
Case 2: $a=1$
$$2^x>x-1=b^x \implies b=1 \implies x=2 \implies (x,y,z)=(2,3,2)$$
Case 3: $x=1$
$$b^x=1-1=0 \implies b=0 \implies z=0$$ However, this is not possible as $x,y,z$ are positive integers.
Thus, for $x,y,z \in \mathbb{N}$, the only solutions are $(x,y,z)=(3,5,6),(2,3,2)$.
If $x,y,z$ are any reals, we can just directly have: $$y=\log_{x}(x^x+z^x)= \frac{\ln(x^x+z^x)}{\ln(x)}$$