Exponential Accuracy!

You can solve it algebraically but would require a calculator and a little casework. Rewrite L.H.S. as $2^X(1+3^X)$ .

Since L.H.S is an increasing function and R.H.S is a decreasing function $\forall X\gt 1$ with no common function values (I found this by several tries in a calculator) and undefined for $X=0$ , so we obviously have to look for values of $X\lt 1,~X\neq 0$ to satisfy the expression as values $\gt 1$ will never satisfy it. For $X\lt 0$ , we can rewrite R.H.S as $\left (-\dfrac{0.39}{X}\right )$ and for $X\in (0,1]$ , we can rewrite it as $\left (\dfrac{0.39}{X}\right )$ .

Using a calculator and checking for $X=1,(-1),(-2),(-3)$ , we can conclude that we should have $X\in [-4,0)\cup (0,1)$ for the approximation to work and the best approximation comes for integer value $X=(-3)$ with the expression value $-0.3\overline{8} \approx -0.39$ . Hence, we have the answer $X=(-3)$

P.S - The approximation can be improved further but then $X$ wouldn't be an integer.

P.P.S - Now that I see it, the best way is indeed using graphical methods.