1.1 Laws of Exponents and Exponential Equations
SL 1.5 SL 1.7
- the words base, exponent (or index) and exponential equation
- the laws of exponents, including zero, negative and fractional exponents
- how to solve an exponential equation by writing both sides with the same base
- how to solve an equation like \(4^x - 3(2^x) - 4 = 0\) with a substitution
Exponential equations
A single bacterium splits into two every 30 minutes. After one splitting there are \(2\) bacteria, after two splittings there are \(2^2 = 4\), and after \(n\) splittings there are \(2^n\).
| Number of splittings | 1 | 2 | 3 | \(\cdots\) | \(n\) |
|---|---|---|---|---|---|
| Number of bacteria | \(2^1 = 2\) | \(2^2 = 4\) | \(2^3 = 8\) | \(\cdots\) | \(2^n\) |
When will there be \(512\) bacteria? We need the value of \(n\) that makes
\[ 2^n = 512 . \]
In the expression \(2^n\), the number \(2\) is the base and \(n\) is the exponent (also called the index; the plural is indices).
An exponential equation is an equation in which the unknown is in the exponent. \(2^n = 512\) is an exponential equation. Since \(512 = 2^9\), the answer is \(n = 9\): after 9 splittings, or \(4.5\) hours.
To solve harder equations like this, we first need the laws of exponents.
The laws of exponents
Definitions — for \(a > 0\) and positive integers \(m\) and \(n\) Not in the formula booklet
\[ a^0 = 1 \qquad a^{-n} = \frac{1}{a^n} \qquad a^{\frac{1}{n}} = \sqrt[n]{a} \qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m} \]
- \(a\): the base (\(a > 0\))
- \(n\): a positive integer (the power, or which root)
- \(m\): a positive integer (the power)
Laws — for \(a, b > 0\) and rational numbers \(m\) and \(n\) Not in the formula booklet
| Same base | Same exponent |
|---|---|
| \(a^m \times a^n = a^{m+n}\) | \(a^n \times b^n = (ab)^n\) |
| \(a^m \div a^n = a^{m-n}\) | \(\dfrac{a^n}{b^n} = \left(\dfrac{a}{b}\right)^n\) |
| \(\left(a^m\right)^n = a^{mn}\) |
- \(a\): a base (\(a > 0\))
- \(m\): an exponent (a rational number)
- \(n\): an exponent (a rational number)
- \(b\): another base (\(b > 0\))
A fractional exponent is easiest to work out root first, then power. For example, \(8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\). Taking the root first keeps the numbers small.
Example 1 Without using a calculator, simplify each of the following.
(a) \(\;27^{\frac{2}{3}} \times 9^{-\frac{1}{2}}\) \(\qquad\) (b) \(\;\dfrac{6^4 \times 2^{-3}}{3^3}\) \(\qquad\) (c) \(\;\left(\sqrt{5}\right)^3 \times \sqrt[4]{5} \div \sqrt[4]{5^3}\)
(a) Take each root first.
\[ 27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9, \qquad 9^{-\frac{1}{2}} = \frac{1}{9^{\frac{1}{2}}} = \frac{1}{3} \]
So \(27^{\frac{2}{3}} \times 9^{-\frac{1}{2}} = 9 \times \frac{1}{3} = 3\).
(b) Write \(6^4\) as \(2^4 \times 3^4\) (same exponent law), then collect each base.
\[ \frac{6^4 \times 2^{-3}}{3^3} = \frac{2^4 \times 3^4 \times 2^{-3}}{3^3} = 2^{4-3} \times 3^{4-3} = 2 \times 3 = 6 \]
(c) Write every term as a power of \(5\).
\[ \left(\sqrt{5}\right)^3 \times \sqrt[4]{5} \div \sqrt[4]{5^3} = 5^{\frac{3}{2}} \times 5^{\frac{1}{4}} \div 5^{\frac{3}{4}} = 5^{\frac{3}{2} + \frac{1}{4} - \frac{3}{4}} = 5^{1} = 5 \]
Example 2 Simplify \(\dfrac{\left(2x^3 y^{-1}\right)^2}{4x^{-2}\,y}\), giving your answer with positive exponents only.
First expand the bracket: every factor inside is raised to the power \(2\).
\[ \left(2x^3 y^{-1}\right)^2 = 2^2\, x^{6}\, y^{-2} = 4x^6 y^{-2} \]
Now divide, subtracting exponents for each base.
\[ \frac{4x^6 y^{-2}}{4x^{-2}\, y} = x^{6-(-2)}\, y^{-2-1} = x^{8} y^{-3} = \frac{x^8}{y^3} \]
Solving by writing both sides with the same base
Equating exponents — for \(a > 0\), \(a \neq 1\) Not in the formula booklet
\[ a^x = a^n \;\Longrightarrow\; x = n \]
- \(a\): the base (\(a > 0\), \(a \neq 1\))
- \(x\): the unknown exponent
- \(n\): a known exponent (a number)
So, to solve an exponential equation, try to write both sides as a power of the same base. Then the exponents must be equal. Useful bases are \(2\), \(3\) and \(5\):
| Base 2 | Base 3 | Base 5 |
|---|---|---|
| \(4 = 2^2,\; 8 = 2^3,\; 16 = 2^4,\; 32 = 2^5\) | \(9 = 3^2,\; 27 = 3^3,\; 81 = 3^4\) | \(25 = 5^2,\; 125 = 5^3\) |
The condition \(a \neq 1\) matters: \(1^3 = 1^7\), but \(3 \neq 7\).
Example 3 Solve each of the following equations.
(a) \(\;9^{x} = 243\) \(\qquad\) (b) \(\;4^{x+1} = 8^{\,1-x}\) \(\qquad\) (c) \(\;\dfrac{25^{x}}{5^{x+3}} = \sqrt{5}\)
(a) \(9 = 3^2\) and \(243 = 3^5\).
\[ \left(3^2\right)^x = 3^5 \;\Longrightarrow\; 3^{2x} = 3^5 \;\Longrightarrow\; 2x = 5 \;\Longrightarrow\; x = \frac{5}{2} \]
(b) \(4 = 2^2\) and \(8 = 2^3\). Keep the brackets around the exponents.
\[ 2^{2(x+1)} = 2^{3(1-x)} \;\Longrightarrow\; 2x + 2 = 3 - 3x \;\Longrightarrow\; 5x = 1 \;\Longrightarrow\; x = \frac{1}{5} \]
(c) Write everything as a power of \(5\): \(25^x = 5^{2x}\) and \(\sqrt{5} = 5^{\frac{1}{2}}\).
\[ \frac{5^{2x}}{5^{x+3}} = 5^{\frac{1}{2}} \;\Longrightarrow\; 5^{2x-(x+3)} = 5^{\frac{1}{2}} \;\Longrightarrow\; x - 3 = \frac{1}{2} \;\Longrightarrow\; x = \frac{7}{2} \]
When there are two unknowns, equating exponents gives two linear equations.
Example 4 Solve the simultaneous equations
\[ 3^{x} \times 9^{y} = 81 \qquad\text{and}\qquad \frac{8^{x}}{2^{y}} = 32 . \]
First equation, base \(3\):
\[ 3^{x} \times 3^{2y} = 3^4 \;\Longrightarrow\; x + 2y = 4 \qquad (1) \]
Second equation, base \(2\):
\[ \frac{2^{3x}}{2^{y}} = 2^5 \;\Longrightarrow\; 3x - y = 5 \qquad (2) \]
From (2), \(y = 3x - 5\). Substitute into (1):
\[ x + 2(3x - 5) = 4 \;\Longrightarrow\; 7x = 14 \;\Longrightarrow\; x = 2 \]
Then \(y = 3(2) - 5 = 1\). So \(x = 2\), \(y = 1\).
Check: \(3^2 \times 9^1 = 81\) ✓ and \(\dfrac{8^2}{2^1} = \dfrac{64}{2} = 32\) ✓
Sometimes the terms cannot be combined directly, but they share a common factor. Take out the smallest power of the base.
Example 5 (a) Show that \(\dfrac{2^{x+3} - 2^{x+1}}{2^{x-1}} = 12\) for all values of \(x\).
(b) Hence solve \(\dfrac{2^{x+3} - 2^{x+1}}{2^{x-1}} = 3 \times 2^{2x}\).
(a) Every term contains \(2^{x-1}\). Write each term as \(2^{x-1}\) times a number:
\[ 2^{x+3} = 2^{x-1} \times 2^4 = 16 \times 2^{x-1}, \qquad 2^{x+1} = 2^{x-1} \times 2^2 = 4 \times 2^{x-1} \]
So
\[ \frac{2^{x+3} - 2^{x+1}}{2^{x-1}} = \frac{2^{x-1}\,(16 - 4)}{2^{x-1}} = 12 . \]
(b) By part (a), the left-hand side is \(12\), so
\[ 12 = 3 \times 2^{2x} \;\Longrightarrow\; 2^{2x} = 4 = 2^2 \;\Longrightarrow\; 2x = 2 \;\Longrightarrow\; x = 1 . \]
Using a substitution
Look at the equation \(9^x - 4(3^x) + 3 = 0\). The two powers have the same base \(3\), because
\[ 9^x = \left(3^2\right)^x = \left(3^x\right)^2 . \]
If we write \(u = 3^x\), the equation becomes \(u^2 - 4u + 3 = 0\) — a quadratic equation in \(u\).
Method — equations like \(a^{2x} + p\,a^x + q = 0\)
- Let \(u = a^x\), so that \(a^{2x} = u^2\).
- Solve the quadratic equation in \(u\).
- Put each value back into \(a^x = u\) and solve for \(x\).
- Reject any value with \(u \le 0\): \(a^x\) is always positive.
Example 6 Use the substitution \(u = 3^x\) to solve \(9^x - 4(3^x) + 3 = 0\).
With \(u = 3^x\) we have \(9^x = u^2\), so
\[ u^2 - 4u + 3 = 0 \;\Longrightarrow\; (u - 1)(u - 3) = 0 \;\Longrightarrow\; u = 1 \;\text{ or }\; u = 3 . \]
Both values are positive, so both give a solution.
\[ 3^x = 1 = 3^0 \;\Longrightarrow\; x = 0, \qquad 3^x = 3 = 3^1 \;\Longrightarrow\; x = 1 . \]
So \(x = 0\) or \(x = 1\).
Example 7 Solve \(4^x + 2^{x+1} - 8 = 0\).
Both terms are powers of \(2\):
\[ 4^x = \left(2^x\right)^2, \qquad 2^{x+1} = 2 \times 2^x . \]
Let \(u = 2^x\):
\[ u^2 + 2u - 8 = 0 \;\Longrightarrow\; (u + 4)(u - 2) = 0 \;\Longrightarrow\; u = -4 \;\text{ or }\; u = 2 . \]
\(2^x = -4\) has no solution, because \(2^x > 0\) for every real \(x\). Reject it.
\[ 2^x = 2 \;\Longrightarrow\; x = 1 . \]
In a Graphs page, enter \(f_1(x) = 4^x + 2^{x+1} - 8\) and use Menu → Analyze Graph → Zero. The graph crosses the \(x\)-axis only once, at \(x = 1\). This confirms that the root \(u = -4\) really gives no solution.
On Paper 1 you have no calculator, so the algebra above is the method you must show.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(2^3 \times 2^4 = 2^{12}\) | \(2^3 \times 2^4 = 2^{7}\) | Multiply powers → add exponents. |
| \(2 \times 3^x = 6^x\) | \(2 \times 3^x\) stays as it is | The bases are different; \(2\) is not raised to the power \(x\). |
| \(2^{-3} = -8\) | \(2^{-3} = \dfrac{1}{8}\) | A negative exponent means a reciprocal, not a negative number. |
| \(2^x + 2^x = 2^{2x}\) | \(2^x + 2^x = 2 \times 2^x = 2^{x+1}\) | Adding terms does not add exponents. |
| \((a + b)^2 = a^2 + b^2\) | \((a + b)^2 = a^2 + 2ab + b^2\) | The exponent laws work for products, not sums. |
| keeping \(3^x = -2\) | rejecting it | \(a^x > 0\), so a negative value of \(u\) gives no solution. |
Exercises
1 Without a calculator, evaluate
(a) \(\;32^{\frac{3}{5}}\) \(\qquad\) (b) \(\;\left(\dfrac{8}{27}\right)^{-\frac{2}{3}}\) \(\qquad\) (c) \(\;0.25^{-\frac{3}{2}}\)
(a) \(32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8\)
(b) A negative exponent turns the fraction upside down: \(\left(\dfrac{8}{27}\right)^{-\frac{2}{3}} = \left(\dfrac{27}{8}\right)^{\frac{2}{3}} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\)
(c) \(0.25 = \dfrac{1}{4}\), so \(0.25^{-\frac{3}{2}} = 4^{\frac{3}{2}} = \left(\sqrt{4}\right)^3 = 2^3 = 8\)
2 Simplify \(\left(3a^2 b^{-1}\right)^3 \times \left(9a^4 b\right)^{-\frac{1}{2}}\), giving your answer with positive exponents only.
\[ \left(3a^2 b^{-1}\right)^3 = 27a^6 b^{-3}, \qquad \left(9a^4 b\right)^{-\frac{1}{2}} = 9^{-\frac{1}{2}} a^{-2} b^{-\frac{1}{2}} = \frac{1}{3}\,a^{-2} b^{-\frac{1}{2}} \]
\[ 27a^6 b^{-3} \times \frac{1}{3}\,a^{-2} b^{-\frac{1}{2}} = 9a^{4} b^{-\frac{7}{2}} = \frac{9a^4}{b^{\frac{7}{2}}} \]
3 Write \(\dfrac{4^3 \times \sqrt{2}}{8^{-1}}\) as a single power of \(2\).
\[ \frac{2^6 \times 2^{\frac{1}{2}}}{2^{-3}} = 2^{6 + \frac{1}{2} - (-3)} = 2^{\frac{19}{2}} \]
4 Solve \(8^{x} = \dfrac{1}{32}\).
\[ 2^{3x} = 2^{-5} \;\Longrightarrow\; 3x = -5 \;\Longrightarrow\; x = -\frac{5}{3} \]
5 Solve \(27^{x+1} = 9^{2x-1}\).
\[ 3^{3(x+1)} = 3^{2(2x-1)} \;\Longrightarrow\; 3x + 3 = 4x - 2 \;\Longrightarrow\; x = 5 \]
6 Solve \(\dfrac{1}{4^{x}} = 2\sqrt{2}\).
Left-hand side: \(\dfrac{1}{4^x} = 4^{-x} = 2^{-2x}\). Right-hand side: \(2\sqrt{2} = 2^1 \times 2^{\frac{1}{2}} = 2^{\frac{3}{2}}\).
\[ 2^{-2x} = 2^{\frac{3}{2}} \;\Longrightarrow\; -2x = \frac{3}{2} \;\Longrightarrow\; x = -\frac{3}{4} \]
7 Solve the simultaneous equations
\[ 2^{x} \times 4^{y} = 64 \qquad\text{and}\qquad \frac{3^{x}}{9^{y}} = \frac{1}{9} . \]
\[ 2^{x + 2y} = 2^6 \;\Longrightarrow\; x + 2y = 6 \qquad (1) \]
\[ 3^{x - 2y} = 3^{-2} \;\Longrightarrow\; x - 2y = -2 \qquad (2) \]
Adding (1) and (2): \(2x = 4\), so \(x = 2\). Then \(2 + 2y = 6\), so \(y = 2\).
8 Show that \(5^{n+2} - 5^{n}\) is a multiple of \(24\) for every positive integer \(n\).
Take out the common factor \(5^n\):
\[ 5^{n+2} - 5^{n} = 5^n \times 5^2 - 5^n = 5^n\,(25 - 1) = 24 \times 5^n . \]
Since \(5^n\) is an integer, \(24 \times 5^n\) is a multiple of \(24\).
9 Solve \(4^{x} - 10\left(2^{x}\right) + 16 = 0\).
Let \(u = 2^x\), so \(4^x = u^2\):
\[ u^2 - 10u + 16 = 0 \;\Longrightarrow\; (u - 2)(u - 8) = 0 \;\Longrightarrow\; u = 2 \;\text{ or }\; u = 8 . \]
\(2^x = 2 \Rightarrow x = 1\), and \(2^x = 8 = 2^3 \Rightarrow x = 3\).
10 Solve \(3^{2x+1} + 8\left(3^{x}\right) - 3 = 0\).
\(3^{2x+1} = 3 \times 3^{2x} = 3\left(3^x\right)^2\). Let \(u = 3^x\):
\[ 3u^2 + 8u - 3 = 0 \;\Longrightarrow\; (3u - 1)(u + 3) = 0 \;\Longrightarrow\; u = \frac{1}{3} \;\text{ or }\; u = -3 . \]
Reject \(u = -3\), because \(3^x > 0\).
\[ 3^x = \frac{1}{3} = 3^{-1} \;\Longrightarrow\; x = -1 \]