If you rewrite the equation as $2^x=-{1\over2}x$, you can see that you are trying to find where the exponential curve $y=2^x$ and the straight line $y=-{1\over2}x$ intersect. Since the exponential curve is increasing (as $x$ increases) while the straight line, which has negative slope, is decreasing, the curves have at most one point of intersection. Thus $(x,y)=(-1,{1\over2})$ is the sole point of intersection, i.e., $x=-1$ is the only solution to the original equation.


Hint: This is equivalent to $2^{x+1}+x=0$. See what happens when you take the derivative...