Week 5

Question 1

5x+ 4 = 19

Answer 1

5x+ 4 = 19

5x+ 4−4 = 19−4

5x= 15

5x/5=15/5

x= 3

Question 2

Answer 2

Question 3

3x+ 9 =−3

Answer 3

Question 4

−2x+ 6 = 26

Answer 4

Question 5

x/5 + 2 = 7

Answer 5

Question 6

2x/3 − 4 = 0

Answer 6

Question 7

6n+ 8 = 44

Answer 7

Question 8

5(3m−1) = 70

Answer 8

Question 9

−18−3h= 5

Answer 9

Question 10

4x+ 7 = 8−x

Answer 10

Question 11

2(4y+ 1) = 5(3−y)

Answer 11

Question 12

7(5−c) = 3(c+ 1)−10

Answer 12

Question 13

Answer 13

Question 14

Answer 14

Question 15

Answer 15

Question 16

The kinetic energy, E, of an object is given by E=1/2mv² where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

Find the kinetic energy, E, of an object that has a mass of 8kg and is travelling at 6m/s.

Answer 16

Question 17

The kinetic energy, E, of an object is given by E=1/2mv² where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

An object has a kinetic energy of 180J and is travelling at 8m/s. Find the mass of the object.

Answer 17

Question 18

The kinetic energy, E, of an object is given by E=1/2mv² where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

Find the velocity of an object if it has a mass of 11.5kg and has a kinetic energy of 81.6J.

Answer 18

Question 19

The surface area, S, of a box with length L, width W and height H is given by

S= 2(L×W) + 2(L×H) + 2(H×W).

Find the surface area of a box with a length of 5cm, width of 3cm and a height of 2.5cm.

Answer 19

Question 20

The surface area, S, of a box with length L, width W and height H is given by

S= 2(L×W) + 2(L×H) + 2(H×W).

Find the height of a box with a surface area of 154.78cm³, a length of 8.5cm and a width of 4.4cm.

Answer 20

Question 21

n²−4n= 0

Answer 21

n²−4n= 0

n(n−4) = 0

n= 0, n= 4

Question 22

3h²−27 = 0

Answer 22

3h²−27 = 0

3(h²−9) = 0

3(h−3)(h+ 3) = 0

h= 3, h=−3

Question 23

k²−13k+ 42 = 0

Answer 23

k²−13k+ 42 = 0

(k−6)(k−7) = 0

k= 6, k= 7

Question 24

3x²+x−2 = 0

Answer 24

3x²+x−2 = 0

(3x−2)(x+ 1) = 0

x=2/3, x=−1

Question 25

2a²= 5a+ 3

Answer 25

2a²= 5a+ 3 2a²−5a−3 = 0 (2a+ 1)(a−3) = 0 a=−1/2, a= 3

Question 26

9y²= 4

Answer 26

9y²= 4 y²=4/9 y=±√4/9 y=±2/3 or y=−2/3, y=2/3

Question 27

2x²+ 4x−6 = 0

Answer 27

2x²+ 4x−6 = 0 2(x²+ 2x−3) = 0 x²+ 2x−3 = 0 (x+ 3)(x−1) = 0 x=−3, x= 1

Question 28

p²= 7p−4

Answer 28

p²= 7p−4 p²−7p+ 4 = 0 This quadratic does not factorise easily. Therefore it is necessary to use the quadratic formula. p=7±√(−7)²−4×1×4**/**2×1 =7±√33 /2

Question 29

2x²−9x+ 5 = 0

Answer 29

Question 30

5x²−3x−1 = 0

Answer 30

Question 31

Twice a number is added to 3 times its square. The result is 16. What was the original number?

Answer 31

Question 32

A soccer ball is kicked such that its height, h, above the ground t seconds after it is kicked is given by h=−t(t−6). How long after it is kicked does it hit the ground?

Answer 32

The equation for the height (h) of the ball t seconds after it has been kicked is given by h=−t(t−6) To find when the ball hits the ground, let h= 0. This gives 0 =−t(t−6) The solutions to this quadratic are t= 0 and t= 6. If we assume that the ball was on the ground at t= 0 and then kicked into the air, the ball will hit the ground six seconds after it has been kicked.

Question 33

log₂(x) = 4

Answer 33

log₂(x) = 4 x= 2⁴ x= 16

Question 34

log₅(x) = 3

Answer 34

log₅(x) = 3 x= 5³ x= 125

Question 35

6 log₄(x) = 12

Answer 35

6 log₄(x) = 12 log₄(x) = 2 x= 4² x= 16

Question 36

log₃(x) + 7 = 10

Answer 36

log₃(x) + 7 = 10 log₃(x) = 3 x= 3³ x= 27

Question 37

log₁₀(5x) = 2

Answer 37

log₁₀(5x) = 2 5x= 10² 5x= 100 x= 20

Question 38

4 log₇(9x) = 8

Answer 38

4 log₇(9x) = 8 log₇(9x) = 2 9x= 7² 9x= 49 x=49/9

Question 39

8 log₂(4x) = 24

Answer 39

8 log₂(4x) = 24 log₂(4x) = 3 4x= 2³ 4x= 8 x= 2

Question 40

5 log₃(6x−1) = 10

Answer 40

5 log₃(6x−1) = 10 log₃(6x−1) = 2 6x−1 = 3² 6x−1 = 9 6x= 10 x=10/6 x=5/3

Question 41

9 log5(3x+ 2) + 7 = 25

Answer 41

9 log5(3x+ 2) + 7 = 25 9 log5(3x+ 2) = 18 log5(3x+ 2) = 2 3x+ 2 = 5² 3x+ 2 = 25 3x= 23 x=23/3

Question 42

3 log4(5x+ 8)−4 = 2

Answer 42

3 log4(5x+ 8)−4 = 2 3 log4(5x+ 8) = 6 log4(5x+ 8) = 2 5x+ 8 = 4² 5x+ 8 = 16 5x= 8 x=8/5

Question 43

2^x= 12

Answer 43

This question could be solved with logarithms of any base. The obvious choices are base 2, base 10 and base e. The solutions for base 2 are shown below. Using base 2 logarithms: 2^x = 12 log₂(2^x) = log₂(12) x \* log₂(2) = log₂(12) x \* 1 = log₂(12) x = log₂(12) However, the question asked us to round our answer to two decimal places. To do this,we will need to use the change of base rule. x= log2(12) =log10(12)/log10(2) = 3.58

Question 44

5^x= 30

Answer 44

5^x= 30 log10(5^x) = log10(30) x × log10(5) = log10(30) x=log10(30)/log10(5) x= 2.11

Question 45

9^x= 58

Answer 45

9^x= 58 log10(9^x) = log10(58) x×log10(9) = log10(58) x=log10(58)/log10(9) x= 1.85

Question 46

e^x= 4

Answer 46

e^x= 4 ln(e^x) = ln(4) x×ln(e) = ln(4) x= 1.39

Question 47

2e^x= 12

Answer 47

2e^x= 12 e^x= 6 ln(e^x) = ln(6) x×ln(e) = ln(6) x= 1.79

Question 48

7e^x= 21

Answer 48

7e^x= 21 e^x= 3 ln(e^x) = ln(3) x×ln(e) = ln(3) x= 1.10

Question 49

5e^x+ 4 = 14

Answer 49

5e^x+ 4 = 14 5e^x= 10 e^x= 2 ln(e^x) = ln(2) x×ln(e) = ln(2) x= 0.69

Question 50

3e^2x+1= 15

Answer 50

3e^2x+1= 15 e^2x+1= 5 ln(e^2x+1) = ln(5) (2x+ 1)×ln(e) = ln(5) 2x+ 1 = ln(5) 2x= ln(5)−1 x=ln(5)−1/2 x= 0.30

Question 51

8e^5x−4= 32

Answer 51

8e^5x−4= 32 e^5x−4= 4 ln(e^5x−4) = ln(4) (5x−4)×ln(e) = ln(4) 5x−4 = ln(4) 5x= ln(4) + 4 x=ln(4)+4/5 x= 1.08

Question 52

6e^9x+7+ 2 = 20

Answer 52

6e^9x+7+ 2 = 20 6e^9x+7= 18 e^9x+7= 3 ln(e^9x+7) = ln(3) (9x+ 7)×ln(e) = ln(3) 9x+ 7 = ln(3) 9x= ln(3)−7 x=ln(3)−7/9 x=−0.66

Question 53

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by m= 3 log₁₀(8t+ 10) Find the mass of the animal when it is born.

Answer 53

m= 3 log₁₀(8t+ 10) To find the mass of the animal at birth, substitute t= 0 into the equation form. (This is because it has been zero weeks since its birth). m= 3 log₁₀(8×0 + 10) = 3 log₁₀(10) = 3 kg

Question 54

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by m= 3 log10(8t+ 10) Find the mass of the animal 10 weeks after it was born

Answer 54

m= 3 log10(8t+ 10) m= 3 log10(8×10 + 10) = 3 log10(90) ≈5.86 kg

Question 55

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by m= 3 log10(8t+ 10) How long until the animal is 8kg?

Answer 55

Question 56

3^2x−4×3^x+ 4 = 0

Answer 56