In short

An exponential function has the form f(x) = a^x where a > 0 and a \neq 1. When a > 1 the function grows without bound; when 0 < a < 1 it decays toward zero. The graph always passes through (0, 1), is always positive, and has the x-axis as a horizontal asymptote. The number e \approx 2.718 is the special base that makes calculus cleanest — the function e^x is its own derivative.

A single bacterium divides into two every 20 minutes. After one division there are 2 bacteria. After two divisions, 4. After three, 8. After ten divisions — about 3 hours and 20 minutes — there are 2^{10} = 1{,}024. After twenty divisions, over a million.

The count at each step is N(n) = 2^n. This is not a polynomial — no polynomial grows this fast. In x^3 the base varies and the exponent is fixed; in 2^x the base is fixed and the exponent varies. That reversal changes everything. Polynomial growth is fast; exponential growth is in a different league entirely. By the time x = 50, the polynomial x^3 gives 125{,}000 while 2^x gives about 10^{15} — a thousand trillion.

This is the core idea: an exponential function is one where the variable sits in the exponent. The fixed base gets multiplied by itself over and over, and that repeated multiplication produces growth (or decay) that no polynomial can match.

The shape of f(x) = a^x

The function f(x) = a^x behaves very differently depending on whether a > 1 or 0 < a < 1.

When a > 1 (exponential growth): The function starts small for large negative x, passes through (0, 1), and climbs steeply for positive x. The larger the base, the steeper the climb.

When 0 < a < 1 (exponential decay): The function starts large for large negative x, passes through (0, 1), and drops toward zero for positive x. Think of it as growth in reverse — moving to the right now means shrinking.

In both cases, f(0) = a^0 = 1. Every exponential function passes through the point (0, 1).

Graphs of 2 to the x and (1/2) to the xTwo exponential curves on the same axes. The curve y equals 2 to the x starts near the x-axis on the left, passes through (0,1), and rises steeply to the right. The curve y equals (1/2) to the x is the mirror image: it starts high on the left, passes through (0,1), and falls toward the x-axis on the right. Both curves stay above the x-axis everywhere. x y 1 2 3 −1 −2 −3 1 2 4 5 3 y = 2ˣ y = (½)ˣ (0, 1)
$y = 2^x$ (growth, dark curve) and $y = \left(\frac{1}{2}\right)^x$ (decay, red curve). Both pass through $(0, 1)$. The growth curve climbs to the right; the decay curve falls. Notice that $\left(\frac{1}{2}\right)^x = 2^{-x}$ — the decay curve is a mirror image of the growth curve across the $y$-axis.

Key properties

Every exponential function f(x) = a^x with a > 0, a \neq 1 shares these properties.

Exponential function

The exponential function with base a is

f(x) = a^x, \quad a > 0,\; a \neq 1

Its properties:

  • Domain: all real numbers (-\infty, \infty).
  • Range: all positive real numbers (0, \infty). The function is never zero and never negative.
  • y-intercept: (0, 1), since a^0 = 1 for every valid base.
  • Horizontal asymptote: the line y = 0 (the x-axis). As x \to -\infty (for a > 1) or x \to +\infty (for 0 < a < 1), the function approaches zero but never reaches it.
  • Monotonicity: strictly increasing when a > 1, strictly decreasing when 0 < a < 1.
  • One-to-one: a^x = a^y implies x = y. Every output corresponds to exactly one input.

The last property — one-to-one — is what makes logarithms possible. Since the exponential function never repeats a value, you can reverse it: given an output, there is exactly one input that produced it. That reverse function is the logarithm.

Why a > 0 and a \neq 1?

If a = 0, then a^x = 0 for x > 0 and is undefined for x < 0 — not useful.

If a < 0, then a^x is undefined for many values of x. For example, (-2)^{1/2} = \sqrt{-2}, which is not a real number. The function would have gaps everywhere.

If a = 1, then a^x = 1^x = 1 for all x — just a constant function, with nothing exponential about it.

So a > 0 and a \neq 1 are the conditions that make a^x a genuine, well-defined, non-trivial function on all of \mathbb{R}.

Comparing exponential growth to polynomial growth

How fast does 2^x grow compared to x^2 or x^{10}? The answer: exponentially faster — eventually. For small x, a polynomial might be larger. But past a certain point, 2^x overtakes any polynomial and stays ahead forever.

x x^2 x^3 2^x
5 25 125 32
10 100 1,000 1,024
20 400 8,000 1,048,576
30 900 27,000 1,073,741,824

At x = 10, 2^x is already comparable to x^3. By x = 30, 2^x is over a billion — dwarfing x^3 = 27{,}000. This is the defining feature of exponential growth: it starts slow (the function is below the polynomial for a while) but eventually dominates any polynomial, no matter how high the degree.

Comparison of x squared, x cubed, and 2 to the xThree curves plotted on the same axes from x equals 0 to x equals 10. The curve y equals x squared grows gently, y equals x cubed grows faster, but y equals 2 to the x starts below both and then overtakes them around x equals 10, shooting upward far more steeply. x y 1 2 4 6 8 9 10
Three growth curves from $x = 0$ to $x = 10$. The polynomial $x^2$ (muted) grows gently. The polynomial $x^3$ (dark) grows faster. But $2^x$ (red) starts below both and then blasts past them — this is the hallmark of exponential growth. Past $x = 10$ the gap only widens.

The number e

Among all possible bases, one stands out: e \approx 2.71828\ldots This irrational number is the natural base of the exponential function, and the function f(x) = e^x has a remarkable property: it is its own derivative. The rate at which e^x grows at any point x equals the value of e^x at that point. No other base has this self-replicating property.

Where does e come from? Consider compound interest. You deposit ₹1 at 100\% annual interest. If compounded once a year, you get ₹2. Compounded twice a year: \left(1 + \frac{1}{2}\right)^2 = ₹2.25. Compounded four times: \left(1 + \frac{1}{4}\right)^4 \approx ₹2.441. Compounded n times:

\left(1 + \frac{1}{n}\right)^n

As n grows without bound — compounding more and more frequently — this expression approaches a limit:

e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828
n \left(1 + \frac{1}{n}\right)^n
1 2
10 2.5937...
100 2.7048...
1,000 2.7169...
10,000 2.7181...
100,000 2.71827...

The value settles toward e. No matter how many times you compound, you can never exceed e — this is the ceiling on continuous compounding.

The limit (1 + 1/n) to the n approaching eA graph showing the values of (1 + 1/n) to the n for increasing values of n. The dots start at 2 for n equals 1 and rise, approaching the horizontal dashed line at y equals e, which is approximately 2.718. The dots get closer and closer to the line but never reach it. n value 1 10 100 1k 10k 100k e 2.718 2.5
$\left(1 + \frac{1}{n}\right)^n$ for increasing $n$. The dots climb toward the dashed line at $e \approx 2.718$ but never cross it. The number $e$ is the limit — the ceiling that continuous compounding approaches.

Why e is special

Three facts that make e the natural choice for the base:

  1. The derivative. \frac{d}{dx} e^x = e^x. The function is its own rate of change. For any other base a, the derivative of a^x is a^x \ln a — an extra constant \ln a appears. Only for a = e does that constant equal 1.

  2. Continuous compounding. If you invest principal P at rate r compounded continuously, the amount after time t is A = Pe^{rt}. The number e emerges naturally from the limit of ever-more-frequent compounding.

  3. The series. e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots — an infinite sum with beautifully clean coefficients. This series converges for every real (and complex) x.

Transformations of exponential functions

The basic curve y = a^x can be shifted and stretched to produce any exponential function you will meet:

Vertical shift of an exponential functionTwo exponential curves. The first is y equals 2 to the x with asymptote at y equals 0. The second is y equals 2 to the x plus 3, shifted up by 3 units, with asymptote at y equals 3. Both curves have the same shape, but the shifted one is higher. x y 1 2 −1 −2 (0, 4) (0, 1) y = 2ˣ y = 2ˣ + 3 y = 0 y = 3
Shifting $y = 2^x$ up by 3 gives $y = 2^x + 3$. The shape is identical — only the position changes. The horizontal asymptote moves from $y = 0$ to $y = 3$, and the $y$-intercept moves from $1$ to $4$.

Interactive: exploring base and growth

Drag the red point along the curve f(x) = 2^x. The readout shows the coordinates. Notice how the function doubles each time x increases by 1 — this constant multiplication factor is the signature of exponential growth.

Interactive exploration of f(x) equals 2 to the xAn interactive graph of f(x) equals 2 to the x. A draggable red point moves along the curve. A readout shows the x and f(x) values. As x increases by 1, f(x) doubles. drag the red point along the curve
Drag the point along $y = 2^x$. Watch the readout: when $x$ goes from $1$ to $2$, $f(x)$ goes from $2$ to $4$ (doubles). From $2$ to $3$, $f(x)$ goes from $4$ to $8$ (doubles again). The constant doubling ratio is what makes the growth exponential.

Example 1: Sketch the graph of $f(x) = 3 \cdot 2^x - 6$ and find its asymptote and intercepts

Step 1. Identify the transformations. Start with the parent function y = 2^x. Multiply by 3: vertical stretch by factor 3. Subtract 6: vertical shift down by 6.

Why: rewriting the function as f(x) = 3 \cdot 2^x - 6 makes the transformations visible — stretch first, then shift.

Step 2. Find the horizontal asymptote. The parent y = 2^x has asymptote y = 0. After the vertical shift of -6, the asymptote moves to y = -6.

Why: as x \to -\infty, 2^x \to 0, so f(x) \to 3(0) - 6 = -6. The function approaches -6 but never reaches it.

Step 3. Find the y-intercept. f(0) = 3 \cdot 2^0 - 6 = 3(1) - 6 = -3. The curve crosses the y-axis at (0, -3).

Why: at x = 0, any base raised to the power 0 is 1, so the y-intercept depends only on the stretch and shift.

Step 4. Find the x-intercept. Set f(x) = 0: 3 \cdot 2^x - 6 = 0, so 3 \cdot 2^x = 6, so 2^x = 2, so x = 1. The curve crosses the x-axis at (1, 0).

Why: the x-intercept exists because the vertical shift pushed the curve below the x-axis. Without the shift, 3 \cdot 2^x would always be positive.

Result. Horizontal asymptote y = -6. y-intercept (0, -3). x-intercept (1, 0). The curve rises steeply for x > 1 and flattens toward -6 for large negative x.

Graph of f(x) equals 3 times 2 to the x minus 6The graph of f(x) equals 3 times 2 to the x minus 6. The curve has a horizontal asymptote at y equals negative 6. It crosses the y-axis at (0, negative 3) and the x-axis at (1, 0). To the right, the curve rises steeply. To the left, it flattens toward y equals negative 6. x y 1 2 3 −1 −2 y = −6 (0, −3) (1, 0)
$f(x) = 3 \cdot 2^x - 6$. The vertical stretch by $3$ makes the curve steeper than $2^x$ alone. The shift down by $6$ moves the asymptote from $y = 0$ to $y = -6$ and creates an $x$-intercept at $(1, 0)$. The curve flattens toward $-6$ on the left and rockets upward on the right.

The asymptote at y = -6 is the floor the function can never break through. No matter how far left you go, the function stays above -6 — the 3 \cdot 2^x term shrinks toward zero but is always positive, keeping the total just above -6.

Example 2: A radioactive substance decays so that the mass remaining after $t$ years is $M(t) = 80 \cdot \left(\frac{1}{2}\right)^{t/5}$. Find the half-life and the mass after $15$ years.

Step 1. Identify the initial mass. M(0) = 80 \cdot \left(\frac{1}{2}\right)^0 = 80 \cdot 1 = 80 grams.

Why: at t = 0 the exponential part is 1, so the initial mass is the coefficient.

Step 2. Find the half-life. The half-life is the time t when M(t) = \frac{80}{2} = 40. Set 80 \cdot \left(\frac{1}{2}\right)^{t/5} = 40, so \left(\frac{1}{2}\right)^{t/5} = \frac{1}{2}, so t/5 = 1, so t = 5 years.

Why: the exponent t/5 becomes 1 when t = 5, meaning the half-life is 5 years. The denominator of the exponent fraction is the half-life.

Step 3. Find the mass after 15 years. M(15) = 80 \cdot \left(\frac{1}{2}\right)^{15/5} = 80 \cdot \left(\frac{1}{2}\right)^3 = 80 \cdot \frac{1}{8} = 10 grams.

Why: 15 years is 3 half-lives. Each half-life halves the mass: 80 \to 40 \to 20 \to 10.

Step 4. Verify the pattern. After 5 years: 40 g. After 10 years: 20 g. After 15 years: 10 g. Each 5-year period cuts the mass in half — exactly what "half-life" means.

Why: the constant ratio (multiply by \frac{1}{2} every 5 years) is the fingerprint of exponential decay, just as constant doubling is the fingerprint of exponential growth.

Result. The half-life is 5 years. After 15 years, 10 grams remain.

Radioactive decay curve with half-life of 5 yearsThe graph of M(t) equals 80 times (1/2) to the t/5. The curve starts at 80 grams at t equals 0, drops to 40 at t equals 5, to 20 at t equals 10, and to 10 at t equals 15. Horizontal dashed lines mark each halving. The curve approaches zero but never reaches it. t (years) M (g) 5 10 15 20 25 80 40 20 10 80 g 40 g 20 g 10 g
$M(t) = 80 \cdot \left(\frac{1}{2}\right)^{t/5}$. Every $5$ years the mass halves: $80 \to 40 \to 20 \to 10 \to \ldots$ The dashed staircase shows each halving step. The curve approaches zero but never reaches it — in principle, some fraction of the substance always remains.

The decay curve tells a vivid physical story: the substance loses half its mass every 5 years, so after 3 half-lives (15 years), only \frac{1}{8} of the original 80 grams — 10 grams — is left. After 10 half-lives (50 years), less than a tenth of a gram would remain. Exponential decay is relentless.

Common confusions

If you understand the shape, the key properties, the role of e, and how to transform and apply exponential functions, you have the essentials — stop here if you like. What follows goes into the algebraic and analytic structure.

The exponential laws, revisited

The familiar laws of exponents — a^m \cdot a^n = a^{m+n}, (a^m)^n = a^{mn}, a^{-n} = 1/a^n — are not just algebraic rules. They are properties of the function f(x) = a^x:

Continuous versus discrete

The bacteria example at the start used integer values of n (each division is a discrete event). But f(x) = 2^x is defined for all real x — including x = 1.5 or x = \pi. What does 2^{1.5} mean? It is 2^{3/2} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2} \approx 2.83. And 2^\pi? It is defined as e^{\pi \ln 2} \approx 8.82. The exponential function smoothly interpolates between integer powers, giving a continuous curve rather than a sequence of dots.

Exponential equations

When you need to solve a^x = b, you take the logarithm of both sides: x = \log_a b = \frac{\ln b}{\ln a}. The one-to-one property guarantees a unique solution (provided b > 0). This is the bridge to the logarithmic function — the inverse of the exponential.

Where this leads next