Solve a system of equations with $3$ unknown powers?

Question

I'm trying to solve this, knowing $X$

$$\begin{cases}1^x*2^y*3^z=X\\x+y+z=13\end{cases}$$

So for example, if $X=2048$ , we have

$$x=2\\y=11\\z=0$$

I barely have memories from high school mathematics, so I'm a bit lost...

Answer 1

$\begin{cases}1^x2^y3^z=X\\x+y+z=13\end{cases}$

$\begin{cases}2^y3^z=X\\x+y+z=13\end{cases}$

$\begin{cases}e^{y\ln2+z\ln3}=X\\x+y+z=13\end{cases}$

$\begin{cases}y\ln2+z\ln3=\ln X+2n\pi i~,~\text{where}~n\in\mathbb{Z}\\x+y+z=13\end{cases}$

Let $x=t$ ,

Then $\begin{cases}y\ln2+z\ln3=\ln X+2n\pi i~,~\text{where}~n\in\mathbb{Z}~......(1)\\t+y+z=13~......(2)\end{cases}$

$(1)-(2)\times\ln3$ :

$y(\ln2-\ln3)=\ln X+2n\pi i-(13-t)\ln3$ , where $n\in\mathbb{Z}$

$y=\dfrac{\ln X+(t-13)\ln3+2n\pi i}{\ln2-\ln3}$ , where $n\in\mathbb{Z}$

$(1)-(2)\times\ln2$ :

$z(\ln3-\ln2)=\ln X+2n\pi i-(13-t)\ln2$ , where $n\in\mathbb{Z}$

$z=\dfrac{\ln X+(t-13)\ln2+2n\pi i}{\ln3-\ln2}$ , where $n\in\mathbb{Z}$

$\therefore\begin{cases}x=t\\y=\dfrac{\ln X+(t-13)\ln3+2n\pi i}{\ln2-\ln3}\\z=\dfrac{\ln X+(t-13)\ln2+2n\pi i}{\ln3-\ln2}\end{cases}$ , where $t\in\mathbb{C}$ and $n\in\mathbb{Z}$