Exponential Functions

Learning goals

  • Put the variable in the exponent, unlike a power function
  • Require b>0b > 0 and b≠1b \ne 1 for the base
  • Read bx+y=bxbyb^{x+y} = b^x b^y as adding input multiplies output
  • Spot exponential data by a positive constant ratio, never 11, at equal steps
  • Describe the graph through (0,1)(0, 1) with asymptote y=0y = 0
  • Tell growth from decay by whether bb exceeds one

The variable belongs in the exponent

You have spent whole chapters on expressions like x2x^2, x3x^3, and general polynomials. In every one of them the variable sits in the base and a fixed number sits in the exponent. An exponential function turns that arrangement around.

A function is exponential when the variable appears in the exponent while the base is a fixed number. Its basic form is

f(x)=bx,f(x) = b^x,

where the constant bb is called the base. A slightly fuller form, f(x)=a⋅bxf(x) = a \cdot b^x, includes a constant multiplier aa that sets the starting value; the plain form bxb^x is just the case a=1a = 1. Throughout the graphing below we take a>0a > 0, so every output stays positive.

The whole idea lives in one comparison. Look at 2x2^x next to x2x^2:

2xversusx2.2^x \qquad \text{versus} \qquad x^2.

They are built from the same two symbols, a 22 and an xx, and beginners read them as if they were the same thing. They are not even close. In x2x^2 the exponent is frozen at 22 and the base xx moves; that is a power function. In 2x2^x the base is frozen at 22 and the exponent xx moves; that is an exponential function. Watch how differently they grow:

xx001122334455
x2x^20011449916162525
2x2^x1122448816163232

The two happen to agree at x=2x = 2 and again at x=4x = 4, but their engines are different, and past x=4x = 4 the exponential pulls away and never looks back. By x=10x = 10 the power function reaches only 102=10010^2 = 100, while the exponential is already at 210=10242^{10} = 1024, more than ten times larger.

The exponential 2 to the x versus the power x squaredy = x squared is a symmetric parabola through the origin; y = 2 to the x is always positive, passes through (0, 1), and rises steeply. The labeled intersection is (2, 4); the curves cross once more near x = -0.77.xy-2-112312468y = 2ˣy = x²(2, 4)(0, 1)(0, 0)
Same symbols, different functions. In y = x² the variable is the base, and the graph is a parabola through the origin. In y = 2ˣ the variable is the exponent, so the graph never touches the axis, passes through (0, 1), and climbs faster and faster. The two meet at (2, 4) as shown and again at x = 4, just past this view; for every x beyond 4, 2ˣ stays above x² for good. (They also cross once more to the left of the y-axis, near x = -0.77, but that comparison is not the point here.)

Worked example 1 Exponential, power, or neither?

Sort each rule by asking one question: where is the variable, the base or the exponent?

f(x)=2xf(x) = 2^x has its variable in the exponent over the fixed base 22, so it is exponential.

g(x)=x2g(x) = x^2 has its variable in the base under the fixed exponent 22, so it is a power function, not exponential.

h(x)=5⋅3xh(x) = 5 \cdot 3^x has its variable in the exponent over the fixed base 33, with a multiplier of 55 in front, so it is exponential.

k(x)=1xk(x) = 1^x looks exponential, but the base is 11, and 11 multiplied by itself any number of times is still 11:

1x=1for every x.1^x = 1 \quad \text{for every } x.

That is a constant, a flat horizontal line, so it is not a genuine exponential function. This is exactly why the base 11 is ruled out, a point the next section settles.

Check your understanding

Which of these is an exponential function?

Answer choices

Which bases make sense

Not every number is allowed as a base. Two restrictions come straight from what bxb^x has to mean if it is going to be a function defined for every input xx.

The base must be positive. Suppose you tried a negative base such as b=−4b = -4. As long as the exponent is a whole number you get an answer, but the sign flips back and forth: (−4)1=−4(-4)^1 = -4, (−4)2=16(-4)^2 = 16, (−4)3=−64(-4)^3 = -64. Worse, a fractional exponent asks for a root of a negative number, and

(−4)1/2=−4(-4)^{1/2} = \sqrt{-4}

is not a real number at all. So a negative base leaves the function undefined at many inputs and lurching in sign at the rest, which is no way to build a smooth curve. The base b=0b = 0 fails too, since 0x0^x is 00 for positive xx and undefined for x≤0x \le 0. Ruling these out, we require b>0b > 0.

The base must not be 11. As the worked example showed, 1x=11^x = 1 for every xx, so the “function” is just the constant 11. It carries none of the growth or decay that makes an exponential interesting, so we set it aside.

Putting the two conditions together, an exponential function f(x)=bxf(x) = b^x uses a base with

b>0andb≠1.b > 0 \quad \text{and} \quad b \ne 1.

Check your understanding

Which of these fails to be a valid base for an exponential function, and why?

Answer choices

From whole-number powers to a smooth curve

The definition bxb^x has to make sense for every real input, not just whole numbers, before we can draw it as an unbroken curve. Everything needed is already in hand from earlier chapters on exponents:

Those rules pin down bxb^x at every rational input. The remaining inputs, the irrational ones like x=2x = \sqrt{2}, get filled in so the curve has no gaps or jumps between the rational points. The result is that bxb^x is defined for every real number xx and traces one smooth, unbroken curve that is positive everywhere.

Here is the curve’s backbone for b=2b = 2:

xx−2-2−1-100112233
2x2^x14\tfrac{1}{4}12\tfrac{1}{2}11224488

Two features of that row are worth naming now, because both trip people up later. The value at x=0x = 0 is 11, not 00. And the values at negative inputs are small positive fractions, not negative numbers.

Worked example 2 Evaluate f(x)=2xf(x) = 2^x at four inputs

Each input calls on one of the rules above.

At a whole-number input, multiply copies of the base:

f(5)=25=32.f(5) = 2^5 = 32.

At zero, every allowed base gives 11:

f(0)=20=1.f(0) = 2^0 = 1.

At a negative input, take the reciprocal of the positive power:

f(−3)=2−3=123=18.f(-3) = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}.

At a fractional input, read the exponent as a root:

f ⁣(12)=21/2=2≈1.41.f\!\left(\tfrac{1}{2}\right) = 2^{1/2} = \sqrt{2} \approx 1.41.

Notice that 2\sqrt 2 lands between f(0)=1f(0) = 1 and f(1)=2f(1) = 2, exactly where the smooth curve says the value for x=12x = \tfrac{1}{2} should fall.

The rule that makes a function exponential

One identity governs the base power bxb^x, and it is the reason every exponential function, including the fuller form a⋅bxa \cdot b^x, behaves the way it does. For the base power, adding inputs turns into multiplying outputs.

Why bx+y=bx⋅byb^{x+y} = b^x \cdot b^y#

Start with whole-number exponents, where you can count factors directly. Write bxb^x as xx copies of bb, and byb^y as yy copies of bb. The product bx⋅byb^x \cdot b^y then lays the two runs of factors end to end, giving x+yx + y copies of bb in all:

bx⋅by=b⋯b⏟x⋅b⋯b⏟y=b⋯b⏟x+y=bx+y.b^x \cdot b^y = \underbrace{b \cdots b}_{x} \cdot \underbrace{b \cdots b}_{y} = \underbrace{b \cdots b}_{x + y} = b^{x + y}.

This whole-number case is not new material on its own: it is the product rule for exponents, the same rule you already used and extended to negative and fractional exponents in earlier chapters. It continues to hold for every real number xx and yy, not just whole numbers, exactly the way the earlier chapters extended it to fractions:

bx+y=bx⋅by.b^{x + y} = b^x \cdot b^y.

Set y=1y = 1 to see what this means step by step. Then bx+1=bx⋅bb^{x+1} = b^x \cdot b, which says that increasing the input by 11 multiplies the output by the base bb. Equal steps added in the exponent produce equal factors multiplied in the value, and that is precisely the exponential behavior.

That one-step consequence, multiply by bb per unit step, carries over to the fuller form a⋅bxa \cdot b^x too, even though the addition identity above is stated for the base power alone. Since a⋅bx+1=b⋅(a⋅bx)a \cdot b^{x+1} = b \cdot \left(a \cdot b^x\right), each step of 11 in the input still multiplies the value of the function by bb, whatever aa is.

This turns hard-looking calculations into easy ones. Since 210=10242^{10} = 1024, you can find nearby powers without multiplying out a long string of 22s. Adding 33 in the exponent multiplies by 23=82^3 = 8, and subtracting 33 divides by 88:

213=210⋅23=1024⋅8=8192,27=210⋅2−3=10248=128.2^{13} = 2^{10} \cdot 2^{3} = 1024 \cdot 8 = 8192, \qquad 2^{7} = 2^{10} \cdot 2^{-3} = \frac{1024}{8} = 128.

Equal steps, constant ratio

The step-by-step reading of the rule, “each unit added to xx multiplies the output by bb,” gives a practical test for spotting exponential behavior in a table. Line up equally spaced inputs and look at what happens between neighbors. A linear function adds the same amount each step, a constant difference. An exponential function multiplies by the same amount each step, a constant ratio.

Compare a linear rule g(x)=3+5xg(x) = 3 + 5x with an exponential rule f(x)=3⋅2xf(x) = 3 \cdot 2^x over the same inputs:

xx0011223344
g(x)=3+5xg(x) = 3 + 5x3388131318182323
f(x)=3⋅2xf(x) = 3 \cdot 2^x3366121224244848

For gg, the jumps are 5,5,5,55, 5, 5, 5, a constant difference, so successive values climb by the same amount. For ff, the jumps are 3,6,12,243, 6, 12, 24, which are not constant at all; instead the ratios 63,126,2412,4824\tfrac{6}{3}, \tfrac{12}{6}, \tfrac{24}{12}, \tfrac{48}{24} are all 22. That is the test: whenever equally spaced inputs produce a constant ratio between the listed outputs, and that ratio is positive and not equal to 11, the data fits an exponential rule, and when those inputs step by 11 that ratio is the base. Those two conditions on the ratio are not extra fine print, they are the base restrictions b>0b > 0 and b≠1b \ne 1 showing up in the data: a ratio of 11 means every output repeats, which is the flat, non-growing case already ruled out, and a ratio can never come out negative, since multiplying by the same positive base again and again never flips the sign. A finite table can only show that the data fits the pattern; it does not by itself pin down what an unknown function does at every other input.

Worked example 3 Recognize exponential data and find the rule

A function’s outputs at x=0,1,2,3x = 0, 1, 2, 3 are 5,15,45,1355, 15, 45, 135. Decide whether it could be exponential, and if so, find its rule.

First test the differences, the linear signature:

15−5=10,45−15=30,135−45=90.15 - 5 = 10, \quad 45 - 15 = 30, \quad 135 - 45 = 90.

The differences are not equal, so the data is not linear. Now test the ratios, the exponential signature:

155=3,4515=3,13545=3.\frac{15}{5} = 3, \quad \frac{45}{15} = 3, \quad \frac{135}{45} = 3.

The ratio is a constant 33, so the data fits an exponential rule with base b=3b = 3. The starting value at x=0x = 0 is 55, which is the multiplier aa, so

f(x)=5⋅3x.f(x) = 5 \cdot 3^x.

Check one point to be sure: f(2)=5⋅32=5⋅9=45f(2) = 5 \cdot 3^2 = 5 \cdot 9 = 45, matching the table.

Check your understanding

A function is exponential, and its outputs at x=0,1,2,3x = 0, 1, 2, 3 are 4,12,36,1084, 12, 36, 108. What is f(4)f(4)?

Answer choices

Growth, decay, and the shape of the graph

Whether an exponential rises or falls is decided entirely by the base.

When b>1b > 1, each step to the right multiplies by a number bigger than 11, so the outputs get larger. The function increases, and the picture is exponential growth.

Graph of the growth exponential y = 2 to the xAn increasing curve, always positive, through (0, 1), with horizontal asymptote y = 0 on the left and a steep rise on the right.xy-3-2-11231248y = 0(0, 1)y = 2ˣb > 1: growth
A growth exponential, base b = 2 greater than 1. The curve passes through (0, 1), stays positive for every input, and rises faster and faster. To the left it flattens toward the horizontal asymptote y = 0 without ever reaching it, so the output is never 0 or negative.

When 0<b<10 < b < 1, each step to the right multiplies by a number smaller than 11, so the outputs shrink toward zero. The function decreases, and the picture is exponential decay. A decay curve is a mirror image of a growth curve, because raising one-half to the xx is the same as raising two to the negative xx: (12)x=2−x\left(\tfrac{1}{2}\right)^x = 2^{-x}.

Graph of the decay exponential y = one-half to the xA decreasing curve, always positive, through (0, 1), with horizontal asymptote y = 0 on the right; the mirror image of the growth curve.xy-3-2-11231248y = 0(0, 1)y = (½)ˣ0 < b < 1: decay
A decay exponential, base b = ½ between 0 and 1. It also passes through (0, 1) and stays positive, but each step to the right multiplies the output by ½, so the curve falls toward the horizontal asymptote y = 0. It is the mirror image of the growth curve, because raising one-half to the x is the same as raising two to the negative x.

Both curves, growth and decay, share the same list of features, and reading them off the picture is the payoff of this lesson:

One base deserves a mention by name. The number e≈2.718e \approx 2.718 is a famous choice of base whose special role shows up in the next lesson, on compound interest. For now, treat it as just another base greater than 11: its graph is a growth curve like the one above.

Check your understanding

Which statement describes the graph of f(x)=5⋅(14)xf(x) = 5 \cdot \left(\tfrac{1}{4}\right)^x?

Answer choices

Worked example 4 Read a decay function

A quantity is modeled by f(x)=80⋅(12)xf(x) = 80 \cdot \left(\tfrac{1}{2}\right)^x. Describe its graph and find f(3)f(3).

Identify the base first. It is 12\tfrac{1}{2}, which is between 00 and 11, so this is exponential decay: the graph decreases. The multiplier in front is 8080, which is the starting value, so the graph crosses the vertical axis at its yy-intercept:

f(0)=80⋅(12)0=80⋅1=80.f(0) = 80 \cdot \left(\tfrac{1}{2}\right)^0 = 80 \cdot 1 = 80.

Because the base is 12\tfrac{1}{2}, each step of 11 to the right multiplies the output by 12\tfrac{1}{2}, that is, it halves. So the values run 80,40,20,10,…80, 40, 20, 10, \ldots, and the third step lands on

f(3)=80⋅(12)3=80⋅18=10.f(3) = 80 \cdot \left(\tfrac{1}{2}\right)^3 = 80 \cdot \frac{1}{8} = 10.

The graph stays above the axis the whole way, sliding toward the horizontal asymptote y=0y = 0 without reaching it.

Common mistakes

Practice

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How the irrational inputs get filled in

The rules above pin down bxb^x at every rational input, b0=1b^0 = 1, b−n=1bnb^{-n} = \dfrac{1}{b^n}, and bm/n=(bn)mb^{m/n} = \left(\sqrt[n]{b}\right)^m. An irrational input like x=2x = \sqrt{2} is not covered by any of those rules directly, but it is boxed in tightly by rational numbers on both sides: 1.4<2<1.51.4 < \sqrt{2} < 1.5, then 1.41<2<1.421.41 < \sqrt{2} < 1.42, then 1.414<2<1.4151.414 < \sqrt{2} < 1.415, and so on, closer every time.

Apply 2x2^x to each pair of bounds. Since 2x2^x only gets bigger as xx gets bigger, the outputs are squeezed the same way:

21.4<22<21.5,21.41<22<21.42,21.414<22<21.415,2^{1.4} < 2^{\sqrt 2} < 2^{1.5}, \qquad 2^{1.41} < 2^{\sqrt 2} < 2^{1.42}, \qquad 2^{1.414} < 2^{\sqrt 2} < 2^{1.415},

and so on. Each of those bounds is a power of 22 with a rational exponent, so every one of them is already a known number. As the rational bounds close in on 2\sqrt{2}, the gap between 2lower bound2^{\text{lower bound}} and 2upper bound2^{\text{upper bound}} shrinks toward nothing, trapping 222^{\sqrt{2}} at exactly one value with no room left for it to be anything else. The same squeezing argument works for every irrational input and every allowed base (for a base below 11, where the function decreases instead of increasing, the bounds simply squeeze from the opposite sides), which is why bxb^x is defined for every real xx, and why the curve has no gaps or jumps anywhere along it.

A bit of history (optional)

In 1965 a trade magazine asked an engineer to predict the coming ten years. Gordon Moore, who helped build the American chip industry, had almost nothing to work from. He had a handful of yearly counts of how many parts one silicon chip could carry.

Read one way the counts looked modest. Read another way they were startling. Each year the number had roughly doubled.

Doubling is not a step, it is a factor, and Moore chose to trust the factor. He wrote that the doubling would carry on. A straight line drawn through his early points promised a few hundred parts by 1975. The doubling promised sixty-five thousand, and the doubling turned out to be the closer guess. He later stretched the period to about two years, but the shape of the claim never moved.

What Moore did is the test in this lesson, run on real data. Line up equally spaced inputs and compare the outputs. Equal differences mean the data fits a linear pattern. A positive constant ratio other than 11 means it fits an exponential one, and that ratio is the base. His counts sat close to a ratio of two, so they were roughly a⋅2xa \cdot 2^x in disguise, for whatever count aa he started from. That is why the straight lines drawn through them fell so absurdly short.