Skip to content

Exponential functions

Algebra · level 7, lesson 7 of 7

A question to start

A lily pad grows in a pond. Each day the patch of lily pads doubles in area. On day 30 the pond is completely covered. On which day was it half covered?

Most people's first guess is day 15. But think backwards: the patch doubles every day, so the day before it was full, it was half full. The answer is day 29. On day 25 it covered only \(\dfrac{1}{32}\) of the pond, about 3 percent: you would hardly notice it, five days before the end.

Doubling is a different kind of growth from adding. This lesson gives it a function, a graph and the algebra to go with it.

The idea

Exponent laws, a reminder

\(b^n\) means \(n\) copies of \(b\) multiplied together; \(b\) is the base, \(n\) the exponent. Counting copies gives the laws (in level 6 you used them with variables):

\[b^m \cdot b^n = b^{m+n}, \qquad \frac{b^m}{b^n} = b^{m-n}, \qquad (b^m)^n = b^{mn}, \qquad (ab)^n = a^n b^n.\]

To keep the laws true for every exponent, we are forced to define \(b^0 = 1\) (since \(\dfrac{b^3}{b^3} = 1\)) and \(b^{-n} = \dfrac{1}{b^n}\) (since \(b^0 \div b^n = b^{-n}\)).

Rational exponents

What should \(8^{1/3}\) mean? If the third law is to hold,

\[\left(8^{1/3}\right)^3 = 8^{\frac13 \times 3} = 8^1 = 8.\]

So \(8^{1/3}\) must be a number whose cube is 8: it is \(\sqrt[3]{8} = 2\). In general, \(b^{1/n} = \sqrt[n]{b}\), the \(n\)th root. And then, using the third law again,

\[b^{m/n} = \left(b^{1/n}\right)^m = \left(\sqrt[n]{b}\right)^m.\]

So \(8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\). You could also square first: \(\sqrt[3]{8^2} = \sqrt[3]{64} = 4\). The same answer, but taking the root first keeps the numbers small. For these definitions we take \(b > 0\), so that every root exists.

Exponential functions

An exponential function has the form

\[y = a \cdot b^x, \qquad b > 0,\; b \ne 1,\]

with the variable in the exponent. (\(b = 1\) is left out because \(1^x\) is always 1, a flat line.) The number \(a\) is the starting value, the value at \(x = 0\), since \(b^0 = 1\).

Compare a linear function and an exponential one in a table:

\(x\) 0 1 2 3 4
\(y = 5 + 3x\) (linear) 5 8 11 14 17
\(y = 5 \cdot 3^x\) (exponential) 5 15 45 135 405

The linear one adds 3 each step: constant first differences. The exponential one multiplies by 3 each step: constant ratios. That is the test for an exponential table: divide each value by the one before; if the answers are all the same, it is exponential, and the answer is \(b\).

The graphs

x y −3 −2 −1 1 2 3 1 2 4 8 y = 2ˣ (growth) y = (½)ˣ (decay) asymptote y = 0

For \(y = 2^x\):

  • At \(x = 0\) it is 1; it doubles with each step right: 2, 4, 8, ...
  • Going left it halves: \(2^{-1} = \tfrac12\), \(2^{-2} = \tfrac14\), ... It gets closer and closer to 0 but never reaches it, because half of a positive number is still positive. The \(x\)-axis, \(y = 0\), is a horizontal asymptote.
  • Domain: all real numbers. Range: \(y > 0\).

\(y = \left(\tfrac12\right)^x\) is the mirror image in the \(y\)-axis, because \(\left(\tfrac12\right)^x = 2^{-x}\). It falls: this is decay.

So: \(b > 1\) gives growth, \(0 < b < 1\) gives decay. Every transformation from lesson 5 works too: for example \(y = 3 \cdot 2^x - 4\) is \(2^x\) stretched by 3 and moved down 4. Its asymptote moves down to \(y = -4\).

Why exponential beats any power in the end

Compare \(x^2\) and \(2^x\). At \(x = 4\) they are equal (16). At \(x = 10\): 100 against 1024. At \(x = 20\): 400 against 1 048 576. Each step adds a little to \(x^2\) but doubles \(2^x\). The lily pond shows the same thing: almost nothing for weeks, then everything at once.

Growth and decay models

  • Doubling every \(D\) units of time: \(N = N_0 \cdot 2^{t/D}\). When \(t = D\), the exponent is 1 and \(N = 2N_0\).
  • Half-life \(H\): \(N = N_0 \left(\tfrac12\right)^{t/H}\).
  • Growing by \(r\) percent per step: \(N = N_0(1 + r)^t\); shrinking by \(r\) percent: \(N = N_0(1 - r)^t\), with \(r\) as a decimal.

Solving by common bases

Exponential functions are one-to-one (each passes the horizontal line test), so if \(b^m = b^n\) then \(m = n\). To solve an equation like \(2^x = 32\), write both sides with the same base: \(32 = 2^5\), so \(x = 5\). Equations that cannot be written with a common base, like \(3^x = 20\), need logarithms: level 8.

Worked examples

1. Rational exponents.

  • \(8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\).
  • \(27^{-2/3} = \dfrac{1}{27^{2/3}} = \dfrac{1}{\left(\sqrt[3]{27}\right)^2} = \dfrac{1}{9}\).
  • \(\left(\dfrac{16}{81}\right)^{3/4} = \left(\dfrac{\sqrt[4]{16}}{\sqrt[4]{81}}\right)^3 = \left(\dfrac23\right)^3 = \dfrac{8}{27}\).

2. Bacteria. A culture starts with 500 bacteria and doubles every 3 hours. How many after 12 hours?

  • \(N = 500 \cdot 2^{t/3}\). At \(t = 12\): \(N = 500 \cdot 2^4 = 500 \times 16 = 8000\).

3. A rover's heat source. The Curiosity and Perseverance rovers on Mars are powered by the heat of plutonium-238, which has a half-life of about 87.7 years. What fraction of the plutonium is left after 20 years? After 175.4 years?

  • Fraction left \(= \left(\tfrac12\right)^{t/87.7}\).
  • After 20 years: \(\left(\tfrac12\right)^{20/87.7} \approx 0.854\), about 85 percent.
  • After 175.4 years, two half-lives: \(\left(\tfrac12\right)^2 = \tfrac14\).
  • (The rovers' electric power falls faster than this, because the parts that turn heat into electricity also wear out.)

4. Common bases.

  • \(2^x = 32 = 2^5\), so \(x = 5\).
  • \(9^{x+1} = 27^x\): write both as powers of 3: \(3^{2(x+1)} = 3^{3x}\), so \(2x + 2 = 3x\), so \(x = 2\). Check: \(9^3 = 729 = 27^2\). ✓
  • \(4^x = \dfrac18\): \(2^{2x} = 2^{-3}\), so \(x = -\dfrac32\).

5. A transformed graph. Describe \(y = 3 \cdot 2^x - 4\).

  • Stretch \(y = 2^x\) vertically by 3, then move down 4. Asymptote \(y = -4\); range \(y > -4\).
  • \(y\)-intercept: \(3 \cdot 1 - 4 = -1\). It crosses the \(x\)-axis where \(2^x = \tfrac43\), between \(x = 0\) and \(x = 1\).

Try it

  1. Evaluate (a) \(25^{3/2}\) (b) \(32^{-2/5}\) (c) \(1000^{2/3}\)
  2. Simplify \(\left(x^{1/2}\right)^4 \cdot x^{-1}\).
  3. A table has \(x\): 0, 1, 2, 3 and \(y\): 5, 15, 45, 135. Show it is exponential and find its equation.
  4. Solve \(3^{x - 1} = 81\).
  5. Solve \(8^x = 4^{x + 1}\).
  6. A car worth 30 000 dollars loses 15 percent of its value each year. Write a model and find its value after 3 years.
  7. Which is bigger at \(x = 10\), \(x^3\) or \(2^x\)? At \(x = 20\)?
  8. For \(y = -\left(\tfrac12\right)^x + 3\), give the asymptote, the range and the \(y\)-intercept.
Answers
  1. (a) \(\left(\sqrt{25}\right)^3 = 5^3 = 125\). (b) \(\dfrac{1}{\left(\sqrt[5]{32}\right)^2} = \dfrac{1}{2^2} = \dfrac14\). (c) \(\left(\sqrt[3]{1000}\right)^2 = 10^2 = 100\).
  2. \(x^{2} \cdot x^{-1} = x^{1} = x\).
  3. Ratios: \(\tfrac{15}{5} = \tfrac{45}{15} = \tfrac{135}{45} = 3\), all the same, so it is exponential with \(b = 3\) and starting value 5: \(y = 5 \cdot 3^x\).
  4. \(81 = 3^4\), so \(x - 1 = 4\), so \(x = 5\).
  5. \(2^{3x} = 2^{2(x + 1)} = 2^{2x + 2}\), so \(3x = 2x + 2\), so \(x = 2\). Check: \(8^2 = 64 = 4^3\). ✓
  6. Each year it keeps 85 percent: \(V = 30\,000 \times 0.85^t\). After 3 years: \(30\,000 \times 0.614125 = 18\,423.75\), about 18 424 dollars.
  7. At 10: \(10^3 = 1000\) and \(2^{10} = 1024\), so \(2^x\) is (just) bigger. At 20: \(20^3 = 8000\) and \(2^{20} = 1\,048\,576\), far bigger.
  8. Reflect \(\left(\tfrac12\right)^x\) in the \(x\)-axis and move up 3. Asymptote \(y = 3\), range \(y < 3\), \(y\)-intercept \(-1 + 3 = 2\).

Common mistake

Multiplying the base by the exponent. \(2^5\) is \(2 \times 2 \times 2 \times 2 \times 2 = 32\), not \(2 \times 5 = 10\). And \(3 \cdot 2^x\) is not \(6^x\): the 3 multiplies after the power is worked out (at \(x = 2\), \(3 \cdot 2^2 = 12\), but \(6^2 = 36\)).

Thinking a negative exponent makes a negative number. \(2^{-3} = \tfrac18\), which is positive. A negative exponent means "one over", not "minus". That is why the graph of \(2^x\) never goes below the \(x\)-axis.

Where it leads