Unit 5, lesson 37

Exponential Functions with Different Bases

How an exponential function multiplies its output by the same factor over every equal input step, how the base decides growth or decay, and how a context narrows the domain.

  • Theory part 73
  • Practice part 74
  • Exponential Functions

A culture of bacteria that triples every hour does not gain the same number of cells each hour. It gains three times as many as the hour before. Change by a fixed factor, rather than by a fixed amount, is the signature of an exponential function, and calculus returns to it again and again in growth, decay and interest.

Key terms

Exponential function and its base

Definition Let aa be a real number with a>0a > 0 and a≠1a \ne 1. The exponential function with base aa is

f(x)=ax,x∈R.f(x) = a^x, \qquad x \in \mathbb{R}.

More generally, a function f(x)=Caxf(x) = C a^x with a constant C≠0C \ne 0 is called an exponential function with base aa and initial value CC. For integer xx the power axa^x is repeated multiplication or division, for rational x=p/qx = p/q with q>0q > 0 it is (aq)p\left(\sqrt[q]{a}\right)^p, and for irrational xx it is the value that the rational powers close in on. If a>1a > 1 the function axa^x is strictly increasing, and if 0<a<10 < a < 1 it is strictly decreasing. The base a=1a = 1 is excluded because 1x=11^x = 1 is the constant function, a separate case. Negative bases are excluded because powers such as (−2)1/2(-2)^{1/2} are not real numbers.

In plain words The variable sits in the exponent and the base is a fixed positive number other than 11. The base is the factor the output is multiplied by each time the input goes up by 11.

Example For f(x)=2xf(x) = 2^x the exact values are

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

Each step of 11 to the right doubles the output. Compare x2x^2, where the variable is the base and the exponent is fixed. That is a power function, not an exponential one.

Growth factor

Definition If a>1a > 1, the base aa of f(x)=Caxf(x) = C a^x is called the growth factor. With C>0C > 0 the function is strictly increasing on R\mathbb{R}, which means x1<x2x_1 < x_2 implies f(x1)<f(x2)f(x_1) < f(x_2). Writing a=1+ra = 1 + r with r>0r > 0, the number rr is the growth rate per unit of input, often given as a percent.

In plain words A growth factor above 11 makes the output bigger by the same multiple at every unit step. A factor of 1.041.04 means 4%4\% growth per step.

Example A culture starts with 500500 cells and triples every hour, so after tt hours it has P(t)=500⋅3tP(t) = 500 \cdot 3^t cells. The exact growth table is

tt (hours)0011223344
P(t)P(t)500500150015004500450013500135004050040500

The hourly gains are 10001000, 30003000, 90009000 and 2700027000. The gains are not constant, but each count is exactly 33 times the one before.

Decay factor

Definition If 0<a<10 < a < 1, the base aa of f(x)=Caxf(x) = C a^x is called the decay factor. With C>0C > 0 the function is strictly decreasing on R\mathbb{R}. Writing a=1−ra = 1 - r with 0<r<10 < r < 1, the number rr is the fraction lost per unit of input. Since

(1a)x=a−x,\left(\frac{1}{a}\right)^x = a^{-x},

the graph of y=(1/a)xy = (1/a)^x is the reflection of the graph of y=axy = a^x across the yy-axis.

In plain words A factor between 00 and 11 keeps the same fraction of what was there at each step, so the output shrinks but never runs out.

Example A patient receives 200200 mg of a medication, and each hour the body removes 30%30\% of what is present. The amount left keeps the fraction 0.70.7 each hour, so

M(t)=200⋅(0.7)t mg.M(t) = 200 \cdot (0.7)^t \text{ mg}.

Then M(1)=140M(1) = 140, M(2)=98M(2) = 98 and M(3)=68.6M(3) = 68.6. The base 0.70.7 is the decay factor and r=0.3r = 0.3 is the hourly loss.

Fixed-factor change

Definition For f(x)=Caxf(x) = C a^x and any fixed increment Δ\Delta, every input xx satisfies

f(x+Δ)f(x)=Cax+ΔCax=aΔ.\frac{f(x + \Delta)}{f(x)} = \frac{C a^{x + \Delta}}{C a^x} = a^{\Delta}.

The ratio depends only on the increment Δ\Delta, not on the starting input xx. Equally spaced inputs therefore give outputs that form a geometric sequence. A linear function has the matching property for differences, since (mx+mΔ+b)−(mx+b)=mΔ(mx + m\Delta + b) - (mx + b) = m\Delta.

In plain words Linear functions add the same amount over equal steps. Exponential functions multiply by the same factor over equal steps, wherever the step starts.

Example Take f(x)=3⋅2xf(x) = 3 \cdot 2^x and steps of Δ=0.5\Delta = 0.5. The factor is 20.5=22^{0.5} = \sqrt{2}. From x=1x = 1 to x=1.5x = 1.5 the output goes from 66 to 626\sqrt{2}, and from x=4x = 4 to x=4.5x = 4.5 it goes from 4848 to 48248\sqrt{2}. Both ratios equal 2≈1.414\sqrt{2} \approx 1.414, even though the second jump in height is much larger.

Positive outputs and the y-intercept

Definition For a>0a > 0 with a≠1a \ne 1,

ax>0for every real x,a0=1.\begin{gathered} a^x > 0 \quad \text{for every real } x, \\[0.4em] a^0 = 1. \end{gathered}

So the range of axa^x is (0,∞)(0, \infty) and its graph crosses the yy-axis at (0,1)(0, 1). For f(x)=Caxf(x) = C a^x the yy-intercept is (0,C)(0, C), and every output has the same sign as CC.

In plain words A positive number raised to any real power stays positive, so the basic graph never touches or crosses the xx-axis. At input 00 any base gives 11, so the initial value CC is where the graph meets the yy-axis.

Example The equations 2x=02^x = 0 and 2x=−42^x = -4 have no real solutions. The function g(x)=5⋅(0.4)xg(x) = 5 \cdot (0.4)^x has yy-intercept (0,5)(0, 5), and g(1)=2g(1) = 2 and g(2)=0.8g(2) = 0.8, all positive.

Horizontal asymptote

Definition A line y=Ly = L is a horizontal asymptote of the graph of ff if f(x)f(x) approaches LL as x→∞x \to \infty or as x→−∞x \to -\infty. For a>1a > 1 the values axa^x approach 00 as x→−∞x \to -\infty, and for 0<a<10 < a < 1 they approach 00 as x→∞x \to \infty. In both cases the line y=0y = 0 is a horizontal asymptote on one side only. For

f(x)=Cax+kf(x) = C a^x + k

the horizontal asymptote is y=ky = k, and the graph never reaches it because Cax≠0C a^x \ne 0.

In plain words On one side the graph flattens out and hugs a horizontal line ever more closely without landing on it. On the other side it rises or falls without bound.

Example For 2x2^x at x=−10x = -10 the value is 2−10=11024≈0.000982^{-10} = \frac{1}{1024} \approx 0.00098, already very close to 00 but still positive. The function h(x)=3⋅2x−4h(x) = 3 \cdot 2^x - 4 has horizontal asymptote y=−4y = -4 and yy-intercept h(0)=3−4=−1h(0) = 3 - 4 = -1, and its range is (−4,∞)(-4, \infty).

Contextual domain

Definition The contextual domain of a model is the set of inputs for which the model is meant to describe the situation. It is a subset of the domain of the formula. For an exponential model the formula accepts every real number, but a stated time window such as 0≤t≤T0 \le t \le T restricts the domain to [0,T][0, T], and the range shrinks to the outputs on that interval. When the quantity is measured only at whole time units, the domain may even be a finite set such as {0,1,2,…,T}\{0, 1, 2, \dots, T\}.

In plain words The formula may work for every input, but the story does not. Negative times before the dose, or times far beyond where the model was tested, are left out.

Example Suppose the medication model M(t)=200⋅(0.7)tM(t) = 200 \cdot (0.7)^t is stated to hold for the first 88 hours. Its contextual domain is [0,8][0, 8]. Because MM is decreasing, the largest value is M(0)=200M(0) = 200 and the smallest is

M(8)=200⋅(0.7)8≈11.53 mg,M(8) = 200 \cdot (0.7)^8 \approx 11.53 \text{ mg},

so on this domain the range is about [11.53,200][11.53, 200]. The asymptote y=0y = 0 still describes the shape of the formula, but the model makes no claim about what happens after 88 hours.

Check yourself

  1. Make an exact table of f(x)=4⋅(32)xf(x) = 4 \cdot \left(\frac{3}{2}\right)^x for x=−1,0,1,2,3x = -1, 0, 1, 2, 3. Is ff increasing or decreasing?
Show the answer

The values are 83\frac{8}{3}, 44, 66, 99 and 272\frac{27}{2}. Each is 32\frac{3}{2} times the one before. The base 32\frac{3}{2} is greater than 11 and C=4>0C = 4 > 0, so ff is increasing.

  1. A car’s value is modeled by V(t)=24000⋅(0.85)tV(t) = 24000 \cdot (0.85)^t dollars for 0≤t≤100 \le t \le 10 years. What fraction of its value does it lose each year, what is it worth after 22 years, and what is the range on this domain?
Show the answer

The decay factor is 0.850.85, so it loses 15%15\% of its value each year. After two years it is worth V(2)=24000⋅0.7225=17340V(2) = 24000 \cdot 0.7225 = 17340 dollars. Since VV is decreasing on [0,10][0, 10], the range runs from V(10)≈4725V(10) \approx 4725 up to V(0)=24000V(0) = 24000 dollars.

  1. For g(x)=7⋅(0.6)xg(x) = 7 \cdot (0.6)^x, find the ratio g(x+2)g(x)\frac{g(x + 2)}{g(x)}. Does it depend on xx?
Show the answer

The ratio is (0.6)2=0.36(0.6)^2 = 0.36 for every real xx. It depends only on the step of 22, not on where the step starts.

How this lesson connects