The base in which $1065 = 13 \cdot 54$ is true How can one find the base $r$ in which $1065 = 13 \cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.

$$r^{3}-5r^{2}-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.

Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
Since $3\cdot4-5$ ends in a zero, your base divides $12-5=7$.  Therefore, your base is either $1$ or $7$.  Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).

To give some details, if $r$ is the base, then, in base $10$, $13_r=1\cdot r+3$ and $54_r=5r+4$.  Therefore, by the distributive law, 
$$
13_r\cdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12\equiv12\pmod{r}.
$$
On the other hand, in base $10$,
$$
1065_r=1\cdot r^3+0\cdot r^2+6r+5\equiv 5\pmod{r}.
$$
Therefore, for equality, it must be that $5\equiv 12\pmod{r}.$.
Let $b$ be the base then

$$
(b+3)(5b+4)=b^3+6b+5\Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$

hence $b = 7$ then

$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5  =b^3+6b+5
$$