52. Solve the equation $2^{x}=-x^{2}+2 x+8$.
Solution. The functions $f(x)=2^{x}$ and $g(x)=-x^{2}+2 x+8$ are defined on the set of all real numbers, where $f(x)=2^{x}$ is an increasing function and has no zeros, while $g(x)=-x^{2}+2 x+8$ (a parabola) is an increasing function on the interval $(-\infty ; 1)$, and a decreasing function on the interval $(1 ;+\infty)$ (the point $x=1$ is the x-coordinate of the vertex) and has zeros at $x=-2$ and $x=4$.

Since $f(-2)=2^{-2}=\frac{1}{4}, g(-2)=0, f(-2)>g(-2) ; f(1)=$ $=2^{1}=2, g(1)=9, f(1)<g(1) ; f(4)=2^{4}=16, g(4)=0, f(4)>g(4)$, the graphs of the functions $f(x)$ and $g(x)$ do not intersect on the interval $(1 ; 4)$ (as they are monotonic and continuous), i.e., the given equation has another root on this interval. [^5]

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-32.jpg?height=905&width=811&top_left_y=233&top_left_x=628)

Fig. 25

If $x \in(-\infty ;-2) \cup(4 ;+\infty)$, then $f(x)>0$ and $g(x)<0$, so there are no more solutions.

Therefore, the given equation has two solutions.

By plotting the graphs of the functions $f(x)$ and $g(x)$ on graph paper in the same coordinate system, we can find these solutions (Fig. 25): $x_{1} \approx-1.9, x_{2} \approx 2.7$