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AdvancedGrades 7–822 min36 / 82

Powers with integer exponents

Natural, zero and negative exponents, all the laws of exponents, the sign of a power of a negative number, comparing powers, rewriting with a common base, and standard form.

Check yourself
In this lesson you will learn
  • Evaluate powers with natural, zero and negative exponents and predict their sign
  • Simplify expressions with the laws of exponents
  • Rewrite powers with a common base to evaluate and compare them
  • Write numbers in standard form and calculate with them

Suppose a bacterium splits in two every 20 minutes. In 5 hours there are 15 divisions, and one bacterium becomes 2 · 2 · … · 2 (15 factors) = 2¹⁵ = 32,768 bacteria. In a whole day there are 72 divisions — far too many factors to write out, yet the short form 2⁷² says it all at a glance. A power is a short way to write a product of equal factors. In this lesson you will learn the laws of exponents, do long calculations in one or two lines, and easily write very large and very small numbers, such as the mass of the Earth or of an electron.

You have already met powers such as x² and a³ in the lesson “Algebraic expressions, monomials and polynomials”, and the sign rules for multiplying negative numbers in “Operations with positive and negative numbers”. Here we take the idea of a power all the way, to zero and negative exponents. The next lesson, “Square roots, nth roots and rational exponents”, goes in the opposite direction: finding roots.

Natural exponents and the sign of a power

Definition
Power with a natural exponent

For a natural number n ≥ 2, aⁿ is the product of n factors, each equal to a. Here a is the base, n is the exponent, and finding the product is called raising to a power. By definition a¹ = a. We read a² as “a squared” and a³ as “a cubed”.

aⁿ = a · a · … · a (n factors)
where:
  • athe base (any number)
  • nthe exponent, a natural number

In the order of operations, powers come before multiplication: 2 · 3² = 2 · 9 = 18, whereas (2 · 3)² = 36 is a different expression.

Evaluating powers with natural exponents

Evaluate: 1) 3⁴; 2) (2/3)³; 3) 0.1³; 4) 5 · 2³ − 4²; 5) 1¹⁰⁰ + 0⁷.

Show solution
1) 3⁴ = 3 · 3 · 3 · 3 = 9 · 9 = 81.
2) Both the numerator and the denominator are raised to the power: (2/3)³ = (2 · 2 · 2)/(3 · 3 · 3) = 8/27.
3) 0.1³ = 0.1 · 0.1 · 0.1 = 0.001, with 1 · 3 = 3 digits after the point.
4) Powers first: 2³ = 8, 4² = 16. Then 5 · 8 − 16 = 40 − 16 = 24.
5) Any power of 1 is 1, and a natural power of 0 is 0: 1 + 0 = 1.

The sign of a power of a negative number can be found without calculating it. Every two negative factors together give a positive product. If the number of factors is even, they pair up and the result is positive; if it is odd, one negative factor is left without a partner and the result is negative.

(−a)²ᵏ = a²ᵏ (−a)²ᵏ⁺¹ = −a²ᵏ⁺¹
where:
  • aany number
  • 2kan even exponent (k is a natural number)
  • 2k + 1an odd exponent

An even power of a negative number is positive, an odd power is negative. So an even power of any number is never negative: x² ≥ 0, x⁴ ≥ 0.

The sign of a power

Evaluate: 1) (−2)⁴ and −2⁴; 2) (−3)³; 3) (−1)¹⁰⁰ + (−1)¹⁰¹; 4) (−0.5)² · (−2)³.

Show solution
1) There is an even number of factors: (−2)⁴ = (−2) · (−2) · (−2) · (−2) = 16. In −2⁴ there are no brackets, so the power applies only to the 2: −2⁴ = −(2⁴) = −16.
2) The exponent is odd, so the result is negative: (−3)³ = −27.
3) (−1)¹⁰⁰ = 1 (even exponent), (−1)¹⁰¹ = −1 (odd exponent); the sum is 0.
4) (−0.5)² = 0.25 and (−2)³ = −8; 0.25 · (−8) = −2.
Interactive
Loading simulation…
Move the n slider. For even n the graph is symmetric about the y-axis and never goes below the x-axis: (−a)ⁿ = aⁿ ≥ 0. For odd n a negative x gives a negative value. For 0 < x < 1 a larger exponent gives a smaller value (0.5² > 0.5³), while for x > 1 it gives a larger one. Come back to n = 0 and negative n after the next section.

The laws of exponents

The laws of exponents are not rules to memorise blindly — all of them come from counting factors. For example, 2³ · 2⁴ = (2 · 2 · 2) · (2 · 2 · 2 · 2): there are 3 + 4 = 7 twos in total, that is 2⁷. When dividing, equal factors in the numerator and the denominator cancel: in 5⁶ / 5⁴ four of the six fives cancel and 5² is left.

aᵐ · aⁿ = aᵐ⁺ⁿ aᵐ / aⁿ = aᵐ⁻ⁿ (a ≠ 0)aᵐ · aⁿ = aᵐ⁺ⁿ aᵐ / aⁿ = aᵐ⁻ⁿ (a ≠ 0)
where:
  • athe common base (a ≠ 0 when dividing)
  • m, nthe exponents

When multiplying powers with the same base, add the exponents; when dividing, subtract the exponent of the denominator from that of the numerator. The base does not change!

Products and quotients

Simplify or evaluate: 1) 2³ · 2⁴; 2) x⁵ · x · x²; 3) 7¹² / 7¹⁰; 4) a⁷ · a³ / a⁸ (a ≠ 0); 5) 3¹⁵ / (3⁶ · 3⁷).

Show solution
1) 2³⁺⁴ = 2⁷ = 128.
2) An x with no written exponent has exponent 1: x⁵⁺¹⁺² = x⁸.
3) 7¹²⁻¹⁰ = 7² = 49 — there is no need to compute 7¹²!
4) a⁷ · a³ = a¹⁰, and a¹⁰ / a⁸ = a².
5) The denominator first: 3⁶ · 3⁷ = 3¹³. Then 3¹⁵⁻¹³ = 3² = 9.
(aᵐ)ⁿ = aᵐⁿ
where:
  • athe base
  • m, nthe exponents, which are multiplied

When raising a power to a power, multiply the exponents: (a³)² = a³ · a³ = a³⁺³ = a⁶. The brackets matter: 2^(3²) = 2⁹ = 512, but (2³)² = 2⁶ = 64.

A power of a power

1) (3²)³; 2) (x⁴)³ · x; 3) (a²)⁵ / a⁷ (a ≠ 0); 4) find the value of (−2³)².

Show solution
1) 3²·³ = 3⁶ = 729.
2) (x⁴)³ = x¹², then x¹² · x = x¹³.
3) (a²)⁵ = a¹⁰, and a¹⁰ / a⁷ = a³.
4) Inside the brackets: −2³ = −8. Then (−8)² = 64 — the even power “eats” the minus sign.
(a · b)ⁿ = aⁿ · bⁿ (a / b)ⁿ = aⁿ / bⁿ (b ≠ 0)(a · b)ⁿ = aⁿ · bⁿ (a / b)ⁿ = aⁿ / bⁿ (b ≠ 0)
where:
  • a, bthe factors (b ≠ 0 in the quotient)
  • nthe common exponent

The power of a product is the product of the powers: (ab)³ = ab · ab · ab = a³b³. We also read the rule from right to left: powers with equal exponents can be combined — 2⁵ · 5⁵ = (2 · 5)⁵.

The power of a product and of a quotient

1) (−2a³b)²; 2) (2/5)³; 3) 2⁵ · 5⁵; 4) 4³ · 25³; 5) 15⁴ / 5⁴.

Show solution
1) Square every factor: (−2)² · (a³)² · b² = 4a⁶b².
2) 2³/5³ = 8/125.
3) The exponents are equal, so multiply the bases: (2 · 5)⁵ = 10⁵ = 100,000.
4) (4 · 25)³ = 100³ = 1,000,000.
5) (15/5)⁴ = 3⁴ = 81.

Zero and negative exponents

The definition of aⁿ makes sense only for natural n: there is no such thing as “zero factors” or “minus three factors”. So we give 2⁰ and 2⁻³ a meaning that keeps all the laws working. Look at the pattern: 2³ = 8, 2² = 4, 2¹ = 2 — each time the exponent drops by 1, the value is halved. Continuing gives 2⁰ = 2/2 = 1, 2⁻¹ = 1/2, 2⁻² = 1/4, 2⁻³ = 1/8. The quotient rule says the same: a⁵ / a⁵ = 1, but also a⁵⁻⁵ = a⁰; a² / a⁵ = 1/a³, but also a²⁻⁵ = a⁻³.

a⁰ = 1 a⁻ⁿ = 1 / aⁿ (a ≠ 0)a⁰ = 1 a⁻ⁿ = 1 / aⁿ (a ≠ 0)
where:
  • aa non-zero base
  • na natural number

A power with exponent zero equals 1, and a negative exponent means taking the reciprocal. The expressions 0⁰ and 0⁻ⁿ are not defined (you cannot divide by zero). With this definition all the laws of exponents hold for any integer exponents.

Values of zero and negative powers

Evaluate: 1) 2⁻³; 2) 10⁻⁴; 3) (−5)⁰ and −5⁰; 4) (−2)⁻³; 5) 3⁻¹ + 6⁻¹.

Show solution
1) 2⁻³ = 1/2³ = 1/8 — a negative exponent does not make the number negative!
2) 10⁻⁴ = 1/10⁴ = 1/10,000 = 0.0001.
3) (−5)⁰ = 1, but −5⁰ = −(5⁰) = −1.
4) (−2)⁻³ = 1/(−2)³ = 1/(−8) = −1/8: the sign comes from the base, not from the sign of the exponent.
5) 3⁻¹ + 6⁻¹ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2.
(a / b)⁻ⁿ = (b / a)ⁿ 1 / a⁻ⁿ = aⁿ(a / b)⁻ⁿ = (b / a)ⁿ 1 / a⁻ⁿ = aⁿ
where:
  • a, bnon-zero numbers
  • na natural number

A negative power of a fraction: flip the fraction and change the sign of the exponent. A power with a negative exponent in the denominator moves to the numerator with a positive exponent.

Negative powers of fractions

Evaluate: 1) (2/3)⁻²; 2) 0.2⁻²; 3) 1/2⁻³; 4) 1.5⁻¹.

Show solution
1) Flip the fraction: (2/3)⁻² = (3/2)² = 9/4.
2) Turn the decimal into a fraction: 0.2 = 1/5, so 0.2⁻² = (1/5)⁻² = 5² = 25.
3) 2⁻³ moves from the denominator to the numerator as 2³: 1/2⁻³ = 2³ = 8.
4) 1.5 = 3/2, so 1.5⁻¹ = 2/3.

Once zero and negative exponents are defined, all five laws work for any integer exponents. Be careful with the signs: open the brackets when subtracting and remember the sign rule when multiplying.

Simplifying with integer exponents

Simplify (the variables are non-zero): 1) 3⁻⁴ · 3⁶; 2) 5⁻² / 5⁻⁴; 3) (2⁻²)⁻³; 4) (a⁻³b²)⁻²; 5) x⁻² · x⁵ / x⁻¹.

Show solution
1) 3⁻⁴⁺⁶ = 3² = 9.
2) 5⁻²⁻⁽⁻⁴⁾ = 5⁻²⁺⁴ = 5² = 25.
3) Multiply the exponents: (−2) · (−3) = 6, so 2⁶ = 64.
4) (a⁻³)⁻² · (b²)⁻² = a⁶b⁻⁴ = a⁶/b⁴.
5) x⁻²⁺⁵ = x³, then x³ / x⁻¹ = x³⁻⁽⁻¹⁾ = x⁴.

Common bases and comparing powers

The laws work only when the bases (or the exponents) are equal. If the bases differ, we split them into prime factors and bring them to a common base: 4 = 2², 8 = 2³, 0.5 = 2⁻¹, 9 = 3², 27 = 3³, 25 = 5², 0.04 = 5⁻². Such problems are common in school-leaving and entrance exams, and this trick solves them in a line or two.

n2ⁿ3ⁿ5ⁿ
1235
24925
3827125
41681625
5322433125
664729
71282187
82566561
9512
101024
Recognising these powers helps you spot a common base quickly: 64 = 2⁶ = 4³ = 8², 81 = 3⁴ = 9², 625 = 5⁴ = 25², 1024 = 2¹⁰ = 4⁵ = 32².
  1. 1
    Split the bases

    Split every base into prime factors and write it as a power: 12 = 2² · 3, 0.125 = 1/8 = 2⁻³.

  2. 2
    Remove the brackets

    Use (aᵐ)ⁿ = aᵐⁿ and (ab)ⁿ = aⁿbⁿ: 12³ = (2² · 3)³ = 2⁶ · 3³.

  3. 3
    Collect equal bases

    For every prime base, add the exponents in the numerator and subtract those in the denominator.

  4. 4
    Compute and check

    Compute the small powers that remain. In such problems the answer is usually an integer or a simple fraction; if you get a huge number, check your work.

Evaluating with a common base

Evaluate: 1) 4¹⁰ / 8⁶; 2) 9⁵ / 27³; 3) 6⁵ / (2⁴ · 3⁶); 4) 0.25⁷ · 4⁸; 5) 12³ · 2⁻⁴ / 18².

Show solution
1) 4¹⁰ = (2²)¹⁰ = 2²⁰, 8⁶ = (2³)⁶ = 2¹⁸; 2²⁰ / 2¹⁸ = 2² = 4.
2) 9⁵ = 3¹⁰, 27³ = 3⁹; 3¹⁰ / 3⁹ = 3.
3) 6⁵ = 2⁵ · 3⁵; 2⁵⁻⁴ · 3⁵⁻⁶ = 2 · 3⁻¹ = 2/3.
4) 0.25 = 1/4 = 4⁻¹: 4⁻⁷ · 4⁸ = 4¹ = 4.
5) 12³ = 2⁶ · 3³, 18² = (2 · 3²)² = 2² · 3⁴. Numerator: 2⁶⁻⁴ · 3³ = 2² · 3³. Quotient: 2²⁻² · 3³⁻⁴ = 3⁻¹ = 1/3.

To compare powers, bring them either to a common base or to a common exponent. Then one of the rules below applies. You can also see this on the graph: in the interactive graph above, change n and watch the values at x = 0.5 and at x = 2.

a > 1, m > n ⇒ aᵐ > aⁿ 0 < a < 1, m > n ⇒ aᵐ < aⁿ 0 < a < b ⇒ aⁿ < bⁿ
where:
  • a, bpositive bases
  • m, ninteger exponents (in the third rule n is a natural number)

If the base is greater than 1, a larger exponent gives a larger power; if the base is between 0 and 1, it is the other way round: 0.5² = 0.25 > 0.5³ = 0.125. With the same natural exponent, the larger positive base gives the larger power.

Comparing powers

Compare: 1) 2³⁰⁰ and 3²⁰⁰; 2) 27⁴ and 9⁷; 3) 0.3⁵ and 0.3⁷; 4) 5⁻³ and 5⁻²; 5) (−2)⁵ and (−3)⁴.

Show solution
1) The bases are prime, so bring the exponents to a common 100: 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰, 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰. 8 < 9, so 2³⁰⁰ < 3²⁰⁰.
2) Use base 3: 27⁴ = 3¹², 9⁷ = 3¹⁴. 3 > 1 and 12 < 14, so 27⁴ < 9⁷.
3) The base is between 0 and 1, so the larger exponent gives the smaller power: 0.3⁵ > 0.3⁷.
4) 5 > 1 and −3 < −2, so 5⁻³ < 5⁻² (1/125 < 1/25).
5) (−2)⁵ = −32 < 0 and (−3)⁴ = 81 > 0, so (−2)⁵ < (−3)⁴ — here the signs decide everything.

Standard form (scientific notation)

Astronomy, physics and chemistry deal with very large and very small numbers. The mass of the Earth is about 5,970,000,000,000,000,000,000,000 kg, and the mass of an electron is 0.000…000911 kg (30 zeros after the point). It is easy to miscount the zeros. Powers of 10 solve the problem: 5.97 · 10²⁴ kg and 9.11 · 10⁻³¹ kg.

Definition
Standard form

Writing a positive number as a · 10ⁿ, where 1 ≤ a < 10 and n is an integer. n is called the order of magnitude: it tells how many “steps of ten” the number lies above or below 1.

a · 10ⁿ, 1 ≤ a < 10
where:
  • aa number with one non-zero digit before the decimal point
  • nthe order of magnitude, an integer

If the decimal point moves k places to the left, n = k (numbers greater than 10); if it moves k places to the right, n = −k (numbers less than 1).

  1. 1
    Place the decimal point

    Move the point so that exactly one non-zero digit stays in front of it: 384,400 → 3.844; 0.00056 → 5.6.

  2. 2
    Count the places

    Count how many places the point moved: 5 places to the left in 384,400, 4 places to the right in 0.00056.

  3. 3
    Choose the sign of the exponent

    If the number is greater than 10, n is positive; if it is less than 1, n is negative: 3.844 · 10⁵, 5.6 · 10⁻⁴.

  4. 4
    Check the condition

    a must always satisfy 1 ≤ a < 10: 38.44 · 10⁴ and 0.56 · 10⁻³ have the right values but are not in standard form.

Writing numbers in standard form

Write in standard form: 1) the average distance from the Earth to the Moon, 384,400 km; 2) 0.00056; 3) 45 · 10⁶; 4) 0.3 · 10⁻²; 5) 7,200,000,000.

Show solution
1) The point moves 5 places to the left: 384,400 km = 3.844 · 10⁵ km.
2) The point moves 4 places to the right: 5.6 · 10⁻⁴.
3) 45 is not in standard form: 45 = 4.5 · 10¹, so 45 · 10⁶ = 4.5 · 10¹⁺⁶ = 4.5 · 10⁷.
4) 0.3 = 3 · 10⁻¹, so 0.3 · 10⁻² = 3 · 10⁻¹⁻² = 3 · 10⁻³.
5) The point moves 9 places to the left: 7.2 · 10⁹.
(a · 10ᵐ) · (b · 10ⁿ) = (a · b) · 10ᵐ⁺ⁿ (a · 10ᵐ) / (b · 10ⁿ) = (a / b) · 10ᵐ⁻ⁿ(a · 10ᵐ) · (b · 10ⁿ) = (a · b) · 10ᵐ⁺ⁿ (a · 10ᵐ) / (b · 10ⁿ) = (a / b) · 10ᵐ⁻ⁿ
where:
  • a, bthe factors in front of the powers of 10
  • m, nthe orders of magnitude

Multiply (divide) a and b separately and the powers of 10 separately. If a · b ≥ 10, move the point one place to the left and add 1 to the order. For addition, first make the orders equal.

Calculating in standard form

1) (3 · 10⁵) · (4 · 10⁻²); 2) (6 · 10⁸) / (2 · 10³); 3) The distance from the Earth to the Sun is about 1.5 · 10⁸ km, and the speed of light is about 3 · 10⁵ km/s. How many seconds does sunlight take to reach the Earth? 4) The mass of the Earth is ≈ 5.97 · 10²⁴ kg and that of the Moon ≈ 7.35 · 10²² kg. About how many times heavier is the Earth? 5) 3.2 · 10⁵ + 4 · 10⁴.

Show solution
1) (3 · 4) · 10⁵⁺⁽⁻²⁾ = 12 · 10³ = 1.2 · 10¹ · 10³ = 1.2 · 10⁴.
2) (6 / 2) · 10⁸⁻³ = 3 · 10⁵.
3) t = s / v = (1.5 · 10⁸) / (3 · 10⁵) = 0.5 · 10³ = 5 · 10² s, that is about 500 s ≈ 8 minutes 20 seconds.
4) (5.97 / 7.35) · 10²⁴⁻²² ≈ 0.812 · 10² ≈ 81 times.
5) Make the orders equal: 4 · 10⁴ = 0.4 · 10⁵; 3.2 · 10⁵ + 0.4 · 10⁵ = 3.6 · 10⁵.
Check yourself: fill in the gap
  1. 1.2⁵ · 2³ / 2⁶ =
  2. 2.(−1)⁷ + (−1)⁸ =
  3. 3.(2/3)⁻² =
  4. 4.4³ · 0.25³ =
  5. 5.0.00072 = 7.2 · 10ⁿ, n =
  6. 6.5⁰ + 5⁻¹ =

Key points

  • aⁿ is the product of n factors equal to a; an even power of a negative number is positive and an odd power is negative, while −aⁿ and (−a)ⁿ are different expressions.
  • aᵐ · aⁿ = aᵐ⁺ⁿ, aᵐ / aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, (ab)ⁿ = aⁿbⁿ, (a/b)ⁿ = aⁿ/bⁿ; there is no such rule for sums.
  • a⁰ = 1 and a⁻ⁿ = 1/aⁿ (a ≠ 0); a negative exponent does not make a number negative, it takes the reciprocal.
  • Split different bases into prime factors and bring them to a common base; to compare, look for a common base or a common exponent.
  • Standard form: a · 10ⁿ with 1 ≤ a < 10; when multiplying, multiply the a’s and add the orders.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
If a ≠ 0, what is a⁰ equal to?