How many m?

If we consider the function $f_m(x) = 3^x + m$ , with domain $\mathbb{R}$ and range $(m,\infty)$ , we are asked to solve the equation $f_m(x) = f_m^{-1}(x)$ . This equation is equivalent to solving $f_m(x) = x$ , or equivalently to solving $g(x) = -m$ , where $g(x) = 3^x - x$ . Thus the equation $f_m(x) = f_m^{-1}(x)$ will have a real solution precisely when $-m$ belongs to the range of $g$ . Now $g$ is differentiable with $g'(x) = 3^x \ln 3 - 1$ , and hence $g$ achieves its minimum when $x = \log_3\big(\tfrac{1}{\ln3}\big)$ , and hence the range of $g$ is $[M,\infty)$ , where $M \; = \; g\big(\log_3\big(\tfrac{1}{\ln 3}\big)\big) \; = \; \frac{1}{\ln3} - \log_3\big(\tfrac{1}{\ln3}\big) \; = \; \frac{1 + \ln(\ln 3)}{\ln 3} \approx 0.9958$ Thus we are interested in values of $m$ such that $-m \ge M$ , or $m \le -M$ . The relevant integer values in $(-15,15)$ are $-1,-2,...,-14$ , and so there are $\boxed{14}$ values of $m$ which give real solutions to the original equation.