Evaluating Powers
Sharpen your skills evaluating positive, negative, and fractional bases, and learn the crucial difference between $-2^4$ and $(-2)^4$.
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What is the value of $(-3)^2$? And what is $-3^2$? They look almost the same on the page, but they give different answers. Can you predict which is positive and which is negative, and why?
Evaluating a power is not just multiplication, it is reading the expression carefully. Brackets, signs and indices interact in ways that catch most students out.
The brackets tell you what the base actually is. In $(-2)^4$, the base is $-2$, the negative is inside the bracket, so it's part of the base. In $-2^4$, the base is just $2$; the negative sits outside and applies after the power is calculated. So $(-2)^4 = 16$ but $-2^4 = -16$.
Know
- $a^1 = a$ and (looking ahead) $a^0 = 1$ for $a \ne 0$
- Negative base raised to even index is positive; odd index keeps sign
- $(a/b)^n = a^n / b^n$
Understand
- The difference between $-a^n$ and $(-a)^n$
- Why brackets are essential for negative bases
- Order of operations including powers
Can Do
- Evaluate $(-3)^4, -3^4$, $(1/2)^3$, $(-2)^5$ accurately
- Apply BIDMAS with powers in mixed expressions
- Predict signs of powers without full calculation
Wrong: "$-3^2 = 9$", treating the negative as part of the base.
Right: $-3^2 = -(3^2) = -9$. The negative is OUTSIDE; the base is just $3$.
Wrong: "$(\tfrac{1}{2})^3 = \tfrac{1}{6}$", multiplying bottom by index.
Right: $(\tfrac{1}{2})^3 = \tfrac{1^3}{2^3} = \tfrac{1}{8}$. Raise top and bottom by the index.
When you multiply two negatives, the result is positive. So $(-2) \times (-2) = +4$. Multiplying another $(-2)$ gives a negative again: $(-2)^3 = -8$.
$(-2)^2 = (-2)(-2) = 4$. $(-2)^3 = (-2)(-2)(-2) = -8$. $(-2)^4 = 16$. $(-2)^5 = -32$. The pattern: even index $\to$ positive; odd index $\to$ negative.
For a fraction, raise both the numerator and denominator. For mixed expressions, follow BIDMAS, powers come before multiply/divide and add/subtract.
$\left(\dfrac{2}{3}\right)^3 = \dfrac{2^3}{3^3} = \dfrac{8}{27}$. In mixed expressions like $5 + 3 \times 2^3$, calculate the power first: $2^3 = 8$, then $3 \times 8 = 24$, then $5 + 24 = 29$.
Watch Me Solve It · 3 examples
- 1(a) $(-3)^4$, negative is inside$(-3) \times (-3) \times (-3) \times (-3)$$= 9 \times 9 = 81$ (even index $\to$ positive).
- 2(b) $-3^4$, negative is outside$-(3^4) = -(81) = -81$
- 3Compare$(-3)^4 = 81$ but $-3^4 = -81$. Brackets change the base.
- 1Raise both top and bottom$\left(\dfrac{3}{4}\right)^2 = \dfrac{3^2}{4^2}$
- 2Evaluate each$3^2 = 9$, $4^2 = 16$
- 3Combine$= \dfrac{9}{16}$
- 1Index first (BIDMAS)$4^2 = 16$
- 2Multiplication next$3 \times 16 = 48$
- 3Addition last$2 + 48 = 50$
Common Pitfalls
Negative bases
- $(-a)^{\text{even}}$ = positive
- $(-a)^{\text{odd}}$ = negative
- $-a^n = -(a^n)$ always
Fraction powers
- $(\tfrac{a}{b})^n = \tfrac{a^n}{b^n}$
- $(\tfrac{1}{2})^3 = \tfrac{1}{8}$
- $(\tfrac{2}{3})^2 = \tfrac{4}{9}$
Specials
- $a^1 = a$
- $1^n = 1$
- $(-1)^{\text{even}} = 1$; $(-1)^{\text{odd}} = -1$
BIDMAS reminder
- Brackets, Indices, Divide/Multiply, Add/Subtract
- Indices come BEFORE multiplication
How are you completing this lesson?
Brain Trainer · 4 problems
Four problems to lock in sign rules and bracket handling.
1 Evaluate $(-5)^2$.
Even index, base is $-5$.$25$2 Evaluate $-5^2$.
Base is just $5$; negative applies after.$-25$3 Evaluate $\left(\dfrac{1}{3}\right)^3$.
Cube both top and bottom.$\dfrac{1}{27}$4 Evaluate $10 - 2^3$.
Index first: $2^3 = 8$. Then $10 - 8$.$2$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Evaluate each, showing one line of working: (a) $(-2)^6$, (b) $-2^6$, (c) $\left(\tfrac{3}{5}\right)^2$.
Q7. Use BIDMAS to evaluate $20 - 3 \times 2^2$.
Q8. Explain, with examples, why $(-a)^n$ and $-a^n$ give the same answer when $n$ is odd, but different answers when $n$ is even. Use $a = 3$ with $n = 2, 3, 4$ to illustrate.
Quick Check
1. B$(-2)^5 = -32$.
2. D$-4^2 = -16$.
3. A$(\tfrac{2}{5})^2 = \tfrac{4}{25}$.
4. C$22$.
5. A$(-1)^{100} = 1$.
Show Your Working Model Answers
Q6 (3 marks): (a) $(-2)^6 = 64$ [1]; (b) $-2^6 = -64$ [1]; (c) $\tfrac{9}{25}$ [1].
Q7 (2 marks): $2^2 = 4$ [1]. $20 - 3 \times 4 = 20 - 12 = 8$ [1].
Q8 (4 marks): $n=2$: $(-3)^2 = 9$, $-3^2 = -9$, different [1]. $n=3$: $(-3)^3 = -27$, $-3^3 = -27$, same [1]. $n=4$: $(-3)^4 = 81$, $-3^4 = -81$, different [1]. Odd indices preserve the negative sign so $(-a)^n = -a^n$; even indices flip negatives to positive, so they differ [1].
The Mystery Power
If $(-x)^n = -x^n$ for a positive number $x$ and a whole-number index $n$, what can you say about $n$? Justify your answer.
Reveal solution
$(-x)^n$ equals $x^n$ when $n$ is even (a positive value) and $-x^n$ when $n$ is odd. For $(-x)^n = -x^n$ we need $n$ odd.
Sign rule
$(-)^{\text{even}} = +$, $(-)^{\text{odd}} = -$
Bracket matters
$(-3)^2 = 9$, $-3^2 = -9$
Fraction power
$(\tfrac{a}{b})^n = \tfrac{a^n}{b^n}$
$1^n$ & $0^n$
$1^n = 1$; $0^n = 0$ for $n \ge 1$
BIDMAS
Indices BEFORE multiply/divide
Memory tip
"No brackets = no negative in the base"
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