Plug in several easy values such as $x=-1, 0, 1, 2, 3$, and $4$. Compare $2^x$ and $x^2$ at each point to see where they are equal and where one is larger than the other.

For $x<0$, how do $2^x$ and $x^2$ each change as $x$ increases toward 0? One is increasing and the other is decreasing; what does that suggest about how many times they can intersect on the negative side?

On $x>0$, you can see they intersect at some obvious points. Think about how quickly an exponential function grows compared to a quadratic as $x$ gets large, and whether that allows more intersections beyond the ones you can find by inspection.

In Desmos, enter y = 2^x on one line and y = x^2 on another line so both graphs are displayed together.

Pan and zoom so you can clearly see the part of the graph from about $x=-3$ to $x=6$ and from $y=-1$ to around $y=20$. Make sure both the left side (negative $x$) and the right side (positive $x$) are visible.

Click or tap where the curves intersect; Desmos will mark each intersection and show its coordinates. Count how many intersection points there are to determine how many real solutions the equation has.

Think of the equation

as asking where the graphs of the two functions

• $y = 2^x$ (an increasing exponential curve)
• $y = x^2$ (a parabola opening upward)

intersect. The number of real solutions is the number of intersection points of these two graphs.

Check small integer values of $x$:

• $x=0$: $2^0 = 1$, but $0^2 = 0$ (not equal).
• $x=1$: $2^1 = 2$, but $1^2 = 1$ (not equal).
• $x=2$: $2^2 = 4$ and $2^2 = 4$, so $x=2$ is a solution.
• $x=3$: $2^3 = 8$, but $3^2 = 9$ (not equal).
• $x=4$: $2^4 = 16$ and $4^2 = 16$, so $x=4$ is another solution.

So there are at least two positive real solutions: $x=2$ and $x=4$.

Define a single function

Real solutions of $2^x = x^2$ are exactly the real zeros of $f(x)$.

First show there is at least one negative solution:

• $f(-1) = 2^{-1} - (-1)^2 = \tfrac12 - 1 = -\tfrac12 < 0$.
• $f(0) = 2^0 - 0^2 = 1 - 0 = 1 > 0$.

Since $f(x)$ is continuous, it must cross zero between $x=-1$ and $x=0$, so there is at least one negative solution.

Now show there cannot be more than one negative solution. For $x<0$:

• $2^x$ is increasing as $x$ increases (moving right).
• $x^2$ is decreasing as $x$ increases, because as you move from a more negative number toward 0, its square gets smaller.

So for any $x_1 < x_2 < 0$:

• $2^{x_1} < 2^{x_2}$ (because $2^x$ is increasing),
• $x_1^2 > x_2^2$ (because $x^2$ is decreasing on the negative side).

Therefore

so $f(x)$ is strictly increasing on $(-\infty,0)$.

A continuous strictly increasing function can cross the $x$-axis at most once. Since $f(x)$ changes sign between $-1$ and $0$, there is exactly one negative real solution.

For $x>0$, take natural logs of both sides of $2^x = x^2$ (this is allowed because both sides are positive):

Rearrange to define

Positive solutions of $2^x = x^2$ correspond to values of $x>0$ where $m(x) = 0$.

Now look at how $m(x)$ changes:

• Its derivative (slope) is

• $m'(x) = 0$ only when $x = \dfrac{2}{\ln 2}$ (this is one specific positive number).
• For $0 < x < \dfrac{2}{\ln 2}$, we have $\dfrac{2}{x} > \ln 2$, so $m'(x) < 0$ and $m(x)$ is decreasing.
• For $x > \dfrac{2}{\ln 2}$, we have $\dfrac{2}{x} < \ln 2$, so $m'(x) > 0$ and $m(x)$ is increasing.

So $m(x)$ decreases, reaches a single minimum, then increases. A continuous function that goes down once then up once can cross the horizontal axis at most twice.

Since $x=2$ and $x=4$ already make $m(x)=0$ (because they are solutions of the original equation), these are the only two positive real solutions.

From the analysis:

• There is exactly one negative real solution (from Step 3).
• There are exactly two positive real solutions, $x=2$ and $x=4$ (from Steps 2 and 4).

Therefore, the equation $2^x = x^2$ has $1 + 2 = 3$ real solutions. The correct answer is 3 (choice C).