Exponents, Surds and Exponential Equations

Exponent and surd skills from Grade 10 and 11 are examined again in Grade 12, usually in the algebra question of Paper 1 and inside calculus and finance work. Questions are short, but each step needs to be shown without a calculator.

Simplifying with exponent laws

$a^{m} \times a^{n} = a^{m + n}$ and $a^{m} \div a^{n} = a^{m - n}$
$(a^{m})^{n} = a^{mn}$ and $(ab)^{n} = a^{n} b^{n}$
$a^{0} = 1$ and $a^{- n} = \frac{1}{a ^{n}}$
$a^{\frac{m}{n}} = n a^{m}$

Prime bases. Write numbers as powers of primes: $8 = 2^{3}$ , $27 = 3^{3}$ .

$8^{\frac{2}{3}} \times 2 7^{- \frac{1}{3}} = 2^{2} \times 3^{- 1} = \frac{4}{3}$

Common factors. When terms are added, factorise:

$\frac{2 ^{x + 2} - 2 ^{x}}{3 \cdot 2 ^{x}} = \frac{2 ^{x} ( 4 - 1 )}{3 \cdot 2 ^{x}} = 1$

You cannot add exponents of terms that are added: $2^{x} + 2^{x} = 2 \cdot 2^{x} = 2^{x + 1}$ , not $2^{2 x}$ .

Simplifying surds

Take out perfect square factors: $ab = a b$ .

$50 - 18 + 8 = 52 - 32 + 22 = 42$

Only like surds can be added or subtracted.

Dividing surds.

$\frac{12 + 27}{3} = \frac{2 3 + 3 3}{3} = 5$

To rationalise a denominator, multiply top and bottom by the surd: $\frac{6}{3} = \frac{6 3}{3} = 23$ .

Solving exponential equations

Same base. Write both sides with the same base, then equate exponents.

$3^{x + 1} + 3^{x} = 36 \Rightarrow 3^{x} (3 + 1) = 36 \Rightarrow 3^{x} = 9 \Rightarrow x = 2$

Quadratic in form. Use $k = 2^{x}$ :

$2^{2 x} - 6 \cdot 2^{x} + 8 = 0 \Rightarrow (k - 2) (k - 4) = 0 \Rightarrow x = 1 or x = 2$

Reject any $k \leq 0$ , because $2^{x}$ is always positive.

Rational exponents. Use $k = x^{\frac{1}{3}}$ in $x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} + 4 = 0$ : $k = 1$ or $k = 4$ , so $x = 1$ or $x = 64$ .

Different bases that will not match. Take logarithms: $5^{x} = 12 \Rightarrow x = lo g_{5} 12 \approx 1, 54$ .

Key terms

Exponent: The power to which a base is raised.
Like surds: Surds with the same number under the root, such as $23$ and $53$ .
Rationalise: Remove a surd from the denominator.
Prime factorisation: Writing a number as a product of primes, such as $72 = 2^{3} \times 3^{2}$ .

Common mistakes

Adding exponents of terms that are added: $2^{x} + 2^{3} \neq = 2^{x + 3}$ .
Writing $a + b = a + b$ .
Keeping a solution where $2^{x}$ equals 0 or a negative number.
Forgetting that $x^{\frac{1}{3}} = 4$ means $x = 4^{3}$ , not $34$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Simplify without a calculator: $8^{\frac{2}{3}} \times 2 7^{- \frac{1}{3}}$

Worked solution

Hidden - try the question first

Solution

$8^{\frac{2}{3}} = (2^{3})^{\frac{2}{3}} = 4$
$2 7^{- \frac{1}{3}} = (3^{3})^{- \frac{1}{3}} = \frac{1}{3}$
$4 \times \frac{1}{3} = \frac{4}{3}$

Question 2

5 marks total

routine

Simplify without a calculator:

(a) $\frac{2 ^{x + 2} - 2 ^{x}}{3 \cdot 2 ^{x}}$ (3)

(b) $50 - 18 + 8$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $\frac{2 ^{x} \cdot 2 ^{2} - 2 ^{x}}{3 \cdot 2 ^{x}}$
$= \frac{2 ^{x} ( 4 - 1 )}{3 \cdot 2 ^{x}}$
$= 1$
(b) $52 - 32 + 22$
$= 42$

Question 3

6 marks total

complex

Solve for $x$ :

(a) $3^{x + 1} + 3^{x} = 36$ (3)

(b) $2^{2 x} - 6 \cdot 2^{x} + 8 = 0$ (3)

Worked solution

Hidden - try the question first

Solution

(a) $\frac{2 ^{x} \cdot 2 ^{2} - 2 ^{x}}{3 \cdot 2 ^{x}}$
$3^{x} = 9 = 3^{2}$
$x = 2$
(b) Let $k = 2^{x}$ : $k^{2} - 6 k + 8 = 0 \Rightarrow (k - 2) (k - 4) = 0$
$2^{x} = 2 \Rightarrow x = 1$
$2^{x} = 4 \Rightarrow x = 2$

Question 4

8 marks total

problem-solving

(a) Solve for $x$ : $x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} + 4 = 0$ (4)

(b) Simplify without a calculator: $\frac{12 + 27}{3}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) Let $\frac{2 ^{x} \cdot 2 ^{2} - 2 ^{x}}{3 \cdot 2 ^{x}}$ : $k^{2} - 5 k + 4 = 0$
$(k - 1) (k - 4) = 0$
$x^{\frac{1}{3}} = 1 \Rightarrow x = 1$
$x^{\frac{1}{3}} = 4 \Rightarrow x = 64$
(b) $\frac{2 3 + 3 3}{3}$
$= 5$
(c) $2^{x} \cdot 5^{x} \cdot 5 = 50 \Rightarrow 1 0^{x} = 10$
$x = 1$

FAQs

Are exponents and surds still examined in Grade 12?

Yes. They are part of the algebra section of Paper 1 and are needed in calculus, where you rewrite roots as powers before differentiating.

How do I solve an exponential equation with different bases?

Try to write both sides with the same prime base. If that is impossible, take logarithms.

Why must 2 to the power x be positive?

A positive base raised to any real power gives a positive number, so equations like $2^{x} = - 3$ have no solution.

Simplifying with exponent laws

Simplifying surds

Solving exponential equations

Key terms

Common mistakes

Practice questions

FAQs

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