Intersection of an exponential function and a line

An intersection of an exponential function and a horizontal line amounts to solving an exponential equation. We find where $y = 2^x$ meets the line $y = 4$ .

Matching the bases

At an intersection $2^x = 4$ . Writing the right side as a power of the same base, $4 = 2^2$ , gives the following.

2^x = 2^2

Once the bases match, the exponents must be equal, so $x = 2$ and the intersection is the single point $(2, 4)$ .

Solving with logarithms

When the right side is not a neat power, use logarithms. Taking $\log_2$ of both sides of $2^x = k$ gives $x = \log_2 k$ . Since a logarithm answers the question of what power of $2$ produces $k$ , this is just the solution of the exponential equation rewritten. Here $x = \log_2 4 = 2$ .

At most one point with a horizontal line

The exponential $y = 2^x$ has a base greater than $1$ , so it is always strictly increasing. Its intersections with a horizontal line $y = k$ are therefore as follows.

$k$	Intersections	Reason
$k > 0$	exactly $1$	monotonic, so it never repeats a height
$k \leq 0$	none	the range is $y > 0$

With a base smaller than $1$ the graph is strictly decreasing instead. The curve $y = \left( \dfrac{1}{2} \right)^x = 2^{-x}$ falls to the right, but it too never repeats a height, so it also meets a horizontal line with $k > 0$ at exactly one point. Monotonicity, whether increasing or decreasing, is what pins the intersection to a single point.

When the line is sloped

For $2^x = x + 1$ , the value $x = 0$ gives $1 = 1$ and $x = 1$ gives $2 = 2$ , so there are two intersections. The exponential curve is convex, so a line can cut it at up to two points. Equations like this generally cannot be solved by algebraic rearrangement at all, and their roots have to be found numerically; a case with such tidy answers is a lucky one.

Applications

Exponential equations appear whenever you ask when a growing quantity reaches a given level. The number of years for a principal to double at $3$ percent compound interest solves $1.03^t = 2$ .

t = \frac{\log 2}{\log 1.03} \approx 23.4

The time for a doubling population of bacteria to reach a given count, and the time for a radioactive substance to decay to half its mass, are equations of the same shape. The large dot on the graph is the intersection $(2, 4)$ .