Powers and roots · applications

Applications: Powers and Roots

10 question types · Secondary 1 · each worked step by step with a figure that follows the steps

O-Level · SAT · GCSE Higher

01

A Colony of Bacteria That Doubles Every Hour

methodWrite Repeated Doubling in Index Notation, Then List the Powers of 2 to Pass a Given Size

A dish holds 5 bacteria cells at the start of an experiment. The number of cells doubles every hour. (a) Write the number of cells after 6 hours in index notation and find its value. (b) After how many whole hours does the number of cells first pass 5000?

550 h5 × 21101 h× 25 × 22202 h× 2after n hours: 5 × 2ncells
Each hour multiplies the number of cells by one more 2: after n hours there are 5 × 2n cells.
Let n be the number of hours that have passed. After 1 hour there are 5 × 2 cells and after 2 hours there are 5 × 2 × 2 = 5 × 22 cells. Each hour multiplies by one more 2, so after n hours there are 5 × 2n cells.
step 1 of 5

Doubling is multiplying by 2. Each hour adds one more factor of 2, so the number of factors of 2 is the number of hours, and index notation counts them. Let n be the number of hours.

  1. Let n be the number of hours that have passed. After 1 hour there are 5 × 2 cells and after 2 hours there are 5 × 2 × 2 = 5 × 22 cells. Each hour multiplies by one more 2, so after n hours there are 5 × 2n cells.
  2. (a) After 6 hours there are 5 × 26 cells. Since 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64, this is 5 × 64 = 320 cells.
  3. For part (b) the number of cells must be more than 5000: 5 × 2n > 5000. Divide both sides by 5: 2n > 1000.
  4. List the powers of 2 near 1000: 28 = 256, 29 = 512 and 210 = 1024. The power 29 is less than 1000, and 210 is the first power of 2 that is more than 1000, so n = 10.
  5. (b) The number of cells first passes 5000 after 10 hours. Check: after 9 hours there are 5 × 512 = 2560 cells, which is less than 5000, and after 10 hours there are 5 × 1024 = 5120 cells.

answer(a) 5 × 26 = 320 cells; (b) 10 hours

techniqueIndex Notation

examsO-Level · GCSE Higher

Common pitfalls

  • Working out 5 × 2 × 6 = 60 for part (a). The index 6 does not multiply the 2. It counts how many times 2 is a factor, so 26 = 64 and not 12.
  • Working out (5 × 2)6 = 106. Only the 2 is repeated each hour. The 5 cells at the start are counted once, so the number is 5 × 26.
02

A Square Plot of Known Area and a Cube-Shaped Tank of Known Volume

methodA Square Root Gives the Side and a Cube Root Gives the Edge, Each Found Between Known Squares and Cubes

A farmer has a square vegetable plot with an area of 196 m2 and a cube-shaped water tank with a volume of 343 m3. (a) Find the length of one side of the plot and the length of fencing needed to go once round it. (b) Find the length of one edge of the tank.

196 m2side ssquare plots × s = 196, so s =√196
The area of a square is its side squared, so s2 = 196 and s = √196.
Let the side of the plot be s m. The area of a square is the side times the side, so s2 = 196 and s = √196. The number −14 also has a square of 196, but a length cannot be negative, so only the positive root is used.
step 1 of 5

The area of a square is its side squared, so the side is the square root of the area. The volume of a cube is its edge cubed, so the edge is the cube root of the volume. Known squares and cubes on either side narrow the search, and the last digit finishes it.

  1. Let the side of the plot be s m. The area of a square is the side times the side, so s2 = 196 and s = √196. The number −14 also has a square of 196, but a length cannot be negative, so only the positive root is used.
  2. 102 = 100 and 202 = 400, so s lies between 10 and 20. A square that ends in 6 comes from a number that ends in 4 or in 6. Try 14: 14 × 14 = 196. So s = 14.
  3. (a) One side of the plot is 14 m. The fencing goes along all four sides, so its length is 4 × 14 = 56 m.
  4. Let the edge of the tank be e m. The volume of a cube is edge × edge × edge, so e3 = 343 and e is the cube root of 343. 53 = 125 and 103 = 1000, so e lies between 5 and 10. Only a number that ends in 7 has a cube that ends in 3, so try 7.
  5. (b) 7 × 7 × 7 = 49 × 7 = 343, so one edge of the tank is 7 m long.

answer(a) side 14 m, fencing 56 m; (b) 7 m

techniqueSquares and Square Roots · Cubes and Cube Roots

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the area by 4, 196 ÷ 4 = 49, to find the side. Dividing by 4 finds a side from the perimeter. The area is the side multiplied by itself, so the side is the square root, 14 m.
  • Dividing the volume by 3 to find the edge. The edge is multiplied by itself three times, not added three times, so the edge is the cube root of 343, which is 7 m.
03

A Square Courtyard Paved with a Large Number of Tiles

methodA Square Root by Prime Factorization: Pair the Prime Factors and Take One from Each Pair

A square courtyard is paved with exactly 1764 identical square tiles. (a) How many tiles lie along each side of the courtyard? (b) The owner wants a larger square courtyard paved with exactly 2304 of the same tiles. How many more tiles will lie along each side?

n × n = 1764, so n =√1764
A square with n tiles along each side holds n2 tiles, so n = √1764.
Let n be the number of tiles along each side. The tiles form n rows of n, so n2 = 1764 and n = √1764.
step 1 of 5

A square with n tiles along each side holds n2 tiles, so n is the square root of the number of tiles. When the number is large, write it as a product of prime factors. Every prime factor of a square number appears an even number of times, so the factors can be put in pairs, and one factor from each pair makes the root.

  1. Let n be the number of tiles along each side. The tiles form n rows of n, so n2 = 1764 and n = √1764.
  2. Write 1764 as a product of prime factors. 1764 = 2 × 882 = 2 × 2 × 441, and 441 = 3 × 147 = 3 × 3 × 49 = 3 × 3 × 7 × 7. So 1764 = 22 × 32 × 72.
  3. (a) Take one factor from each pair, which halves every index: √1764 = 2 × 3 × 7 = 42. There are 42 tiles along each side. Check: 42 × 42 = 1600 + 160 + 4 = 1764.
  4. Do the same for 2304. Dividing by 2 again and again gives 1152, 576, 288, 144, 72, 36, 18 and 9, which is eight divisions, and 9 = 32. So 2304 = 28 × 32, and halving every index gives √2304 = 24 × 3 = 16 × 3 = 48.
  5. (b) The larger courtyard has 48 tiles along each side, which is 48 − 42 = 6 more tiles on each side. Check: 48 × 48 = 2304.

answer(a) 42 tiles; (b) 6 more tiles on each side, 48 in all

techniqueRoots by Prime Factorization · Squares and Square Roots

examsO-Level

Common pitfalls

  • Halving the number instead of the indices, 1764 ÷ 2 = 882. A square root is not a half. In the prime factorization it is every index that is halved: 22 × 32 × 72 becomes 2 × 3 × 7.
  • Subtracting the numbers of tiles first and taking the root of the difference, √2304 − 1764 = √540. The root of a difference is not the difference of the roots. Find each side first, then subtract: 48 − 42 = 6.
04

Computer Storage Counted in Powers of 2

methodWith the Same Base, Subtract the Indices to Divide and Add the Indices to Multiply

Computer storage is counted in powers of 2: 1 megabyte (MB) is 220 bytes and 1 gigabyte (GB) is 230 bytes. A memory card holds 235 bytes and one photograph takes up 222 bytes. (a) How many photographs fit on the card? (b) A second card has twice the capacity of the first. Write its capacity in bytes as a power of 2, and find its capacity in gigabytes.

1 MB = 220bytes1 GB = 230bytescard = 235bytesphoto = 222bytesphotographs = 235/ 222
The number of photographs is the capacity of the card divided by the size of one photograph.
The number of photographs is the capacity of the card divided by the size of one photograph: 235 ÷ 222.
step 1 of 5

Every quantity here is a power of the same base, 2. The laws of indices turn a division of powers into a subtraction of indices, am ÷ an = am − n, and a multiplication into an addition, am × an = am + n.

  1. The number of photographs is the capacity of the card divided by the size of one photograph: 235 ÷ 222.
  2. Both powers have the base 2, so subtract the indices: 235 ÷ 222 = 235 − 22 = 213.
  3. (a) 213 = 210 × 23 = 1024 × 8 = 8192, so 8192 photographs fit on the card.
  4. The second card holds 2 × 235 bytes. The number 2 on its own is 21, and multiplying powers of the same base adds the indices: 21 × 235 = 21 + 35 = 236 bytes.
  5. (b) One gigabyte is 230 bytes, so the second card holds 236 ÷ 230 = 236 − 30 = 26 = 64 GB. Check: the first card holds 235 ÷ 230 = 25 = 32 GB, and twice 32 is 64.

answer(a) 213 = 8192 photographs; (b) 236 bytes, which is 64 GB

techniqueLaws of Indices · Index Notation

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the indices, 35 ÷ 22, or dividing the bases to get 113. When powers of the same base are divided, the base stays and the indices are subtracted: 235 − 22 = 213.
  • Writing 2 × 235 as 435 or as 270. Doubling adds just one more factor of 2, so the index goes up by 1: 2 × 235 = 236.
05

Light Halved by Each Pane of Tinted Glass

methodA Negative Index Means One Over the Power

Light of intensity 4800 lux shines on a stack of panes of tinted glass. Each pane lets through exactly half of the light that reaches it. (a) Write the fraction of the light that is left after 5 panes as a power of 2 and as a fraction, and find the intensity of the light behind the 5th pane. (b) How many panes are needed before the intensity first falls below 100 lux?

2010 panes2−11/21 panehalf2−21/42 paneshalfafter n panes: 2−nof the light
One pane leaves 12 = 2−1 of the light, so n panes leave (2−1)n = 2−n of it.
Let n be the number of panes. One pane leaves 12 = 2−1 of the light. Each further pane multiplies by 2−1 again, so after n panes the fraction left is (2−1)n = 2−n.
step 1 of 5

Halving is multiplying by 12, and 12 is 2−1. After n panes the light has been multiplied by 2−1 a total of n times. Let n be the number of panes.

  1. Let n be the number of panes. One pane leaves 12 = 2−1 of the light. Each further pane multiplies by 2−1 again, so after n panes the fraction left is (2−1)n = 2−n.
  2. (a) After 5 panes the fraction left is 2−5. A negative index means one over the power, so 2−5 = 125 = 132.
  3. The intensity behind the 5th pane is 4800 × 132 = 4800 ÷ 32 = 150 lux.
  4. For part (b) the intensity must be less than 100 lux: 4800 × 2−n < 100, which is 48002n < 100. Multiply both sides by 2n and divide both sides by 100: 48 < 2n.
  5. 25 = 32 is not more than 48, and 26 = 64 is. (b) 6 panes are needed. Check: 4800 ÷ 64 = 75 lux, which is below 100 lux, while 5 panes still leave 150 lux.

answer(a) 2−5 = 132, which leaves 150 lux; (b) 6 panes

techniqueZero and Negative Indices · The Power of a Power Rule

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Reading 2−5 as a negative number such as −32 or −10. A negative index does not make the value negative. It means one over the power, so 2−5 = 132, a small positive number.
  • Taking away half of the first intensity for every pane, 4800 − 5 × 2400. Each pane halves the light that reaches it, not the light at the start, so the intensities are 2400, 1200, 600, 300 and 150 lux.
06

The Distances of Two Planets from the Sun in Standard Form

methodDivide the Numbers and the Powers of Ten Separately, and Make the Powers of Ten Equal Before Adding

The Earth is about 1.5 × 108 km from the Sun and Neptune is about 4.5 × 109 km from the Sun. (a) How many times as far from the Sun as the Earth is Neptune? (b) At one time the Earth and Neptune are in a straight line with the Sun, on opposite sides of it. Find the distance between the two planets then, in standard form.

Neptune 4.5 × 109km, Earth 1.5 × 108km(4.5 × 109) divided by (1.5 × 108)
"How many times as far" is a division: (4.5 × 109) ÷ (1.5 × 108).
The question "how many times as far" is a division: (4.5 × 109) ÷ (1.5 × 108).
step 1 of 5

A number in standard form is A × 10n with 1 ≤ A < 10. To divide two such numbers, divide the values of A and subtract the indices of the powers of ten. To add them, first write both with the same power of ten, because only then are the values of A counting the same unit.

  1. The question "how many times as far" is a division: (4.5 × 109) ÷ (1.5 × 108).
  2. Divide the numbers and the powers of ten separately. 4.5 ÷ 1.5 = 3, and 109 ÷ 108 = 109 − 8 = 101.
  3. (a) 3 × 101 = 30, so Neptune is 30 times as far from the Sun as the Earth is.
  4. With the Sun between them, the distance between the planets is the sum of the two distances. To add, the powers of ten must be equal. Write the Earth's distance with 109: the power of ten becomes 10 times as large, so the number in front becomes 10 times as small, 1.5 × 108 = 0.15 × 109.
  5. (b) 4.5 × 109 + 0.15 × 109 = (4.5 + 0.15) × 109 = 4.65 × 109 km. Since 4.65 lies between 1 and 10, this is already in standard form.

answer(a) 30 times as far; (b) 4.65 × 109 km

techniqueCalculating in Standard Form · Standard Form

examsO-Level · GCSE Higher

Common pitfalls

  • Adding the numbers in front without making the powers equal, (4.5 + 1.5) × 109 = 6 × 109. The 1.5 counts lots of 108 and the 4.5 counts lots of 109, so the Earth's distance must first be written as 0.15 × 109.
  • Adding the indices in part (b) to get 1017. Indices are added when powers are multiplied. In an addition with equal powers of ten, the power stays as 109 and only the numbers in front are added.
07

Kinetic Energy from a Formula with a Squared Speed

methodSubstitute into the Formula, Then Work Out the Power Before Multiplying

The kinetic energy E joules of an object of mass m kg moving at a speed of v meters per second is given by the formula E = 12mv2. A cyclist and her bicycle have a total mass of 80 kg. (a) Find the kinetic energy when she rides at 5 meters per second. (b) Find the kinetic energy when she rides at 10 meters per second, and say how many times as large it is.

E = 1/2 × m × v2E = 1/2 × 80 × 52
Substitute m = 80 and v = 5: E = 12 × 80 × 52.
Substitute m = 80 and v = 5 into the formula: E = 12 × 80 × 52.
step 1 of 5

In 12mv2 the index 2 belongs to v only. The order of operations says that a power is worked out before a multiplication, so the speed is squared first and the result is then multiplied by 12 and by m.

  1. Substitute m = 80 and v = 5 into the formula: E = 12 × 80 × 52.
  2. Work out the power first: 52 = 25. The formula becomes E = 12 × 80 × 25.
  3. (a) 12 × 80 = 40 and 40 × 25 = 1000, so the kinetic energy is 1000 joules.
  4. For a speed of 10 meters per second, E = 12 × 80 × 102 = 12 × 80 × 100 = 40 × 100 = 4000 joules.
  5. (b) The kinetic energy is 4000 joules, and 4000 ÷ 1000 = 4, so it is 4 times as large. The speed was doubled and the speed is squared in the formula, so the energy is multiplied by 22 = 4.

answer(a) 1000 joules; (b) 4000 joules, which is 4 times as large

techniquePowers in the Order of Operations

examsGCSE Higher

Common pitfalls

  • Multiplying first and squaring the result, (12 × 80 × 5)2 = 2002 = 40000. The index 2 belongs to v only, and a power is worked out before a multiplication.
  • Saying that twice the speed gives twice the energy, 2000 joules. The speed is squared in the formula, so doubling the speed multiplies the energy by 22 = 4.
08

A Cube-Shaped Box with Eight Times the Volume of Another

methodA Fractional Index Is a Root, and It Applies to Each Factor of a Product

A cube of volume V cm3 has an edge of length V1/3 cm. A small cube-shaped gift box has a volume of 216 cm3. A large cube-shaped gift box has 8 times the volume of the small box. (a) Find the length of one edge of the small box. (b) Find the length of one edge of the large box, and say how many times as long it is.

216 cm3edge ?edge = V1/3, the cube root of V
The index 13 is the cube root, so the edge of the small box is 2161/3 cm.
The index 13 is the cube root: (V1/3)3 = V13 × 3 = V1 = V, so V1/3 is the number whose cube is V. The edge of the small box is 2161/3 cm.
step 1 of 5

The index 13 means the cube root, because cubing V1/3 multiplies the indices and gives V1. A fractional index applies to each factor of a product, (ab)1/3 = a1/3 × b1/3, so the edge of the large box can be found without working out its volume.

  1. The index 13 is the cube root: (V1/3)3 = V13 × 3 = V1 = V, so V1/3 is the number whose cube is V. The edge of the small box is 2161/3 cm.
  2. (a) 53 = 125 and 63 = 6 × 6 × 6 = 216, so 2161/3 = 6. One edge of the small box is 6 cm long.
  3. The large box has a volume of 8 × 216 cm3, so its edge is (8 × 216)1/3 cm. The index applies to each factor of the product: (8 × 216)1/3 = 81/3 × 2161/3.
  4. 81/3 = 2, because 23 = 8. So the edge of the large box is 2 × 6 = 12 cm.
  5. (b) One edge of the large box is 12 cm long, which is 12 ÷ 6 = 2 times as long. Check: 123 = 1728 and 8 × 216 = 1728.

answer(a) 6 cm; (b) 12 cm, which is 2 times as long

techniqueFractional Indices · Cubes and Cube Roots

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Reading 2161/3 as 216 ÷ 3 = 72. The index 13 is not a division by 3. It is the cube root, the number whose cube is 216, which is 6.
  • Saying that 8 times the volume gives 8 times the edge, 48 cm. The volume is the edge cubed, so the edge is multiplied by 81/3 = 2 only. A cube with an edge of 48 cm would have 512 times the volume.
09

An Exact Side of a Right-Angled Garden Bed and an Exact Perimeter

methodPythagoras' Theorem, Simplify the Surd, Then Add the Like Surds

A gardener marks out a flower bed in the shape of a right-angled triangle. The two sides that meet at the right angle are 2 m and 4 m long. In a right-angled triangle the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras' theorem). (a) Find the exact length of the longest side, as a surd in its simplest form. (b) She marks out a mirror image of the bed on the other side of the 4 m side, so that the two beds together make one large triangular bed. Find the exact perimeter of the large bed.

2 m4 mcc2= 22+ 42= 4 + 16 = 20
By Pythagoras' theorem, c2 = 22 + 42 = 20.
Let the longest side be c m. By Pythagoras' theorem, c2 = 22 + 42 = 4 + 16 = 20.
step 1 of 5

Pythagoras' theorem gives the square of the longest side. Its square root is a surd, a root that does not come out as an exact whole number or fraction (it is also called a radical), and it is simplified by taking out a square factor. Surds with the same number under the root are like surds, and they are added like like terms in algebra.

  1. Let the longest side be c m. By Pythagoras' theorem, c2 = 22 + 42 = 4 + 16 = 20.
  2. So c = √20, taking the positive root because c is a length. Take out the largest square factor: 20 = 4 × 5, so √20 = √4 × √5 = 2√5.
  3. (a) The longest side is exactly 2√5 m. The number √5 is irrational, so its decimal, 2.236…, never ends, and 4.47 m is only an approximation. The surd is the exact length.
  4. The large bed is a triangle with two sloping sides of 2√5 m each and a base of 2 + 2 = 4 m. The 4 m side that the two beds share is inside the large bed, so it is not part of the perimeter.
  5. (b) The perimeter is 4 + 2√5 + 2√5 = (4 + 4√5) m. The two like surds add to 4√5, just as 2x + 2x = 4x. The whole number 4 and the surd 4√5 are not like terms, so they cannot be combined any further.

answer(a) 2√5 m; (b) (4 + 4√5) m

techniqueAdding and Multiplying Radicals · Radicals · Rational and Irrational Numbers

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the root of each square separately, √4 + 16 = 2 + 4 = 6. The root of a sum is not the sum of the roots. Add the squares first, then take the root: √20 = 2√5, which is about 4.47.
  • Writing 4 + 4√5 = 8√5. Only like surds can be added. The 4 has no √5, so the exact perimeter stays as 4 + 4√5.
10

The Ratio of a Square's Side to Its Diagonal

methodRationalize the Denominator: Multiply the Numerator and the Denominator by the Surd

A diagonal of a square cuts the square into two right-angled triangles, so by Pythagoras' theorem the square of the diagonal is equal to the sum of the squares of two sides. (a) Write the ratio of the side of a square to its diagonal as a fraction with a rational denominator. (b) A square mirror has a diagonal of 60 cm. Find the exact length of one side of the mirror.

dssd2= s2+ s2= 2s2, so d = s√2
By Pythagoras' theorem, d2 = s2 + s2 = 2s2, so d = s√2.
Let the side of the square be s and the diagonal be d. By Pythagoras' theorem, d2 = s2 + s2 = 2s2, so d = √2s2 = s√2.
step 1 of 5

Let the side of the square be s and its diagonal be d. Pythagoras' theorem gives d in terms of s, and the ratio of the side to the diagonal then has a surd in the denominator. Multiplying the numerator and the denominator by that surd makes the denominator a whole number without changing the value of the fraction.

  1. Let the side of the square be s and the diagonal be d. By Pythagoras' theorem, d2 = s2 + s2 = 2s2, so d = √2s2 = s√2.
  2. The ratio of the side to the diagonal is ss√2. Divide the numerator and the denominator by s: the ratio is 1√2.
  3. (a) Rationalize the denominator: multiply the numerator and the denominator by √2. 1 × √2√2 × √2 = √22, because √2 × √2 = 2.
  4. The side of any square is √22 of its diagonal. For the mirror, the side is 60 × √22 = 30√2 cm.
  5. (b) One side of the mirror is exactly 30√2 cm long, which is about 42.4 cm. Check: (30√2)2 = 900 × 2 = 1800, and two such squares add up to 3600 = 602.

answer(a) √22; (b) 30√2 cm

techniqueRationalizing the Denominator · Radicals

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying only the denominator by √2 and writing 12. That changes the value of the fraction. The numerator and the denominator must both be multiplied by √2, which gives √22.
  • Taking the side as half of the diagonal, 30 cm. A square with a side of 30 cm has a diagonal of 30√2 cm, which is about 42.4 cm and not 60 cm. The side is √22 of the diagonal, not 12 of it.
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