Pure: Exponentials and Logarithms - Worksheets, Questions and Revision

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A-Level · Pure Mathematics

P6 Pure: Exponentials and Logarithms

EDEXCEL 9MA0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question tests index and logarithm notation. Do not use a calculator.

(a)Write log₅(125) = 3 in index form.(1)

(b)Simplify log₃(45) - log₃(5), giving your answer as an integer.(2)

(c)Solve 4^x = 64, showing your method clearly.(2)

(d)Express 2log_a(3) + log_a(5) as a single logarithm.(2)

(Total for Question 1 is 7 marks)

This question tests basic index and logarithm equations.

(a)Solve 2^x = 32, without using a calculator.(2)

(b)Solve 3^x-1 = 9, without using a calculator.(2)

(c)Express 3ln(2) - ln(4) as a single logarithm in its simplest exact form.(2)

(d)Solve log₅(2x + 3) = 2.(3)

(Total for Question 2 is 9 marks)

Solve the following equations, using logarithms where necessary and giving each answer to 3 significant figures.

(a)Solve 5^x = 30.(3)

(b)Solve 3^2x+1 = 50.(4)

(c)Solve 7^x = 4^x+1.(4)

(Total for Question 3 is 11 marks)

Solve the following equations involving natural logarithms.

(a)Solve e^3x = 20, giving your answer to 3 significant figures.(3)

(b)Solve ln(2x - 1) = 4, giving your answer to 3 significant figures.(3)

(c)Solve ln(x) + ln(x + 3) = ln(10), stating why one of the solutions of the resulting quadratic must be rejected.(4)

(Total for Question 4 is 10 marks)

A radioactive isotope decays according to the model m = m0 * e^-λ*t, where m0 grams is the initial mass, t is the time in years and λ is a positive constant. A sample initially has mass 80 g. After 50 years the mass has reduced to 65 g.

(a)Show that λ = (1/50) * ln(80/65), giving the value of λ to 3 significant figures.(3)

(b)Find the half-life of the isotope, giving your answer to 3 significant figures.(3)

(c)Find the time taken for the mass to reduce to 10% of its initial value, giving your answer to the nearest year.(3)

(Total for Question 5 is 9 marks)

A microbiologist called Priya is growing a culture of bacteria. The number of bacteria, N, present t hours after the culture is first observed is modelled by N = 200 * e^0.15*t.

(a)State the number of bacteria present when the culture is first observed.(1)

(b)Find the number of bacteria present after 8 hours, giving your answer to the nearest whole number.(2)

(c)Find the time taken for the number of bacteria to reach 1000, giving your answer to 3 significant figures.(3)

(d)Show that dN/dt = 30*e^0.15*t, and hence find the rate of increase of the number of bacteria when t = 8, giving your answer to 3 significant figures.(4)

(e)Comment on a limitation of this model for large values of t.(1)

(Total for Question 6 is 11 marks)

Consider the equation 2^2x - 9(2^x) + 8 = 0.

(a)Using the substitution y = 2^x, show that the equation can be written as y² - 9y + 8 = 0.(2)

(b)Hence solve 2^2x - 9(2^x) + 8 = 0, giving your answers as exact values of x.(5)

(Total for Question 7 is 7 marks)

A curve C has equation y = ln(x + 2) - 3, for x > -2.

(a)State the equation of the asymptote of C.(1)

(b)Find the coordinates of the point where C crosses the y-axis, giving the y-coordinate to 3 significant figures.(2)

(c)Find the exact coordinates of the point where C crosses the x-axis.(3)

(d)Sketch the graph of C, showing clearly the asymptote from part (a) and the axis-crossing points found in parts (b) and (c).(3)

(Total for Question 8 is 9 marks)

This question tests manipulation of logarithms.

(a)Given that log₂(x) = p, express log₂(4x³) in terms of p.(3)

(b)Solve the equation log₃(x + 5) - log₃(x - 1) = 2, stating clearly any restriction on the value of x.(5)

(Total for Question 9 is 8 marks)

Solve the simultaneous equations log₂(x) + log₂(y) = 5 and log₂(x) - log₂(y) = 1, giving x and y as exact values.

(a)Solve the simultaneous equations, giving x and y as exact values.(6)

(Total for Question 10 is 6 marks)

Rebecca collects data on two variables x and y, which she believes are related by the equation y = A*xⁿ, where A and n are constants. She plots log₁₀(y) against log₁₀(x) and obtains a straight line passing through the points (0, 1.20) and (2, 3.60).

(a)Show that taking logarithms of y = A*xⁿ gives log₁₀(y) = n*log₁₀(x) + log₁₀(A), and explain why this represents a linear relationship between log₁₀(y) and log₁₀(x).(2)

(b)Find the value of n.(2)

(c)Find the value of A, giving your answer to 3 significant figures.(3)

(d)Hence write down the equation connecting y and x.(1)

(Total for Question 11 is 8 marks)

Solve the equation (log₂(x))² - 5(log₂(x)) + 6 = 0, giving your answers as exact values of x.

(a)Solve the equation, giving your answers as exact values of x.(6)

(Total for Question 12 is 6 marks)

Given that x = ln(3 + 2*√2), show that e^x + e^-x = 6.

(a)Show that e^x + e^-x = 6.(5)

(Total for Question 13 is 5 marks)

Jamal owns a rare coin. The value, V pounds, of the coin t years after purchase is modelled by V = 250*e^k*t, where k is a positive constant. The value of the coin doubles every 12 years.

(a)Show that k = ln(2)/12, giving the value of k to 3 significant figures.(4)

(b)Find the value of the coin after 20 years, giving your answer to the nearest pound.(3)

(c)Find the number of complete years after purchase before the coin's value first exceeds 1000 pounds, justifying your answer carefully.(4)

(Total for Question 14 is 11 marks)

Mark scheme · P6 Pure: Exponentials and Logarithms

Question 1

(a) B1 5³ = 125 stated correctly
(a) Answer: 5³ = 125
(b) M1 combines using log₃(45) - log₃(5) = log₃(45/5) = log₃(9)
(b) A1 2 cao
(b) Answer: 2
(c) M1 writes 64 as 4³ (or takes logs of both sides)
(c) A1 x = 3 cao
(c) Answer: x = 3
(d) M1 rewrites 2log_a(3) as log_a(9)
(d) A1 log_a(45) oe cao
(d) Answer: log_a(45)

Question 2

(a) M1 writes 32 as 2⁵
(a) A1 x = 5 cao
(a) Answer: x = 5
(b) M1 writes 9 as 3² and equates exponents x-1 = 2
(b) A1 x = 3 cao
(b) Answer: x = 3
(c) M1 writes 3ln(2) as ln(8) and combines to ln(8/4)
(c) A1 ln(2) cao
(c) Answer: ln(2)
(d) M1 writes 2x + 3 = 5²
(d) M1 rearranges to isolate x
(d) A1 x = 11 cao
(d) Answer: x = 11

Question 3

(a) M1 takes logs (or ln) of both sides: x ln(5) = ln(30)
(a) M1 rearranges to x = ln(30)/ln(5)
(a) A1 x = awrt 2.11
(a) Answer: x = awrt 2.11
(b) M1 takes logs of both sides: (2x+1)ln(3) = ln(50)
(b) M1 rearranges to 2x + 1 = ln(50)/ln(3)
(b) M1 correctly isolates x
(b) A1 x = awrt 1.28
(b) Answer: x = awrt 1.28
(c) M1 takes logs of both sides: x ln(7) = (x+1) ln(4)
(c) M1 expands the right-hand side: x ln(7) = x ln(4) + ln(4)
(c) M1 collects x terms and factorises: x(ln(7) - ln(4)) = ln(4)
(c) A1 x = awrt 2.48
(c) Answer: x = awrt 2.48

Question 4

(a) M1 takes ln of both sides: 3x = ln(20)
(a) M1 rearranges x = ln(20)/3
(a) A1 x = awrt 0.999
(a) Answer: x = awrt 0.999
(b) M1 writes 2x - 1 = e⁴
(b) M1 rearranges to x = (e⁴ + 1)/2
(b) A1 x = awrt 27.8
(b) Answer: x = awrt 27.8
(c) M1 combines logs: ln(x(x+3)) = ln(10)
(c) M1 forms the quadratic x² + 3x - 10 = 0
(c) A1 factorises to give x = 2 or x = -5
(c) B1 rejects x = -5 since ln(x) is undefined for x ≤ 0; x = 2 cao
(c) Answer: x = 2

Question 5

(a) M1 forms the equation 65 = 80 * e^-50*λ
(a) M1 takes ln of both sides and rearranges to λ = (1/50)ln(80/65), cso
(a) A1 λ = awrt 0.00415
(a) Answer: λ = (1/50)ln(80/65) = awrt 0.00415
(b) M1 sets 0.5 = e^-λ*T using the half-life condition
(b) M1 rearranges to T = ln(2)/λ
(b) A1 T = awrt 167 (years), using the unrounded value of λ
(b) Answer: T = awrt 167 years
(c) M1 sets 0.1 = e^-λ*t
(c) M1 rearranges to t = ln(10)/λ
(c) A1 t = awrt 554 (years), using the unrounded value of λ
(c) Answer: t = awrt 554 years

Question 6

(a) B1 N = 200
(a) Answer: 200
(b) M1 substitutes t = 8 into the model
(b) A1 N = awrt 664
(b) Answer: N = awrt 664
(c) M1 sets 1000 = 200*e^0.15*t and simplifies to 5 = e^0.15*t
(c) M1 rearranges to t = ln(5)/0.15
(c) A1 t = awrt 10.7 (hours)
(c) Answer: t = awrt 10.7 hours
(d) M1 differentiates N with respect to t using the chain rule
(d) A1 dN/dt = 200*0.15*e^0.15*t = 30*e^0.15*t, cso
(d) M1 substitutes t = 8 into dN/dt
(d) A1 rate = awrt 99.6 bacteria per hour
(d) Answer: dN/dt = 30e^0.15t; rate at t=8 is awrt 99.6 per hour
(e) B1 any valid limitation, e.g. the model predicts unlimited growth, but in reality growth would be restricted by limited nutrients, space or competition
(e) Answer: the model assumes unrestricted exponential growth, which is unrealistic over long periods as resources are finite

Question 7

(a) B1 recognises 2^2x = (2^x)² = y²
(a) B1 substitutes correctly to obtain y² - 9y + 8 = 0, cso, answer printed
(a) Answer: y² - 9y + 8 = 0 (shown)
(b) M1 factorises y² - 9y + 8 as (y-1)(y-8)
(b) A1 y = 1 and y = 8
(b) M1 converts back to x using 2^x = 1 and 2^x = 8
(b) A1 x = 0 (exact)
(b) A1 x = 3 (exact) cao
(b) Answer: x = 0 or x = 3

Question 8

(a) B1 x = -2
(a) Answer: x = -2
(b) M1 substitutes x = 0 into the equation of C
(b) A1 (0, awrt -2.31)
(b) Answer: (0, -2.31)
(c) M1 sets y = 0: ln(x+2) = 3
(c) M1 rearranges to x + 2 = e³
(c) A1 (e³ - 2, 0) exact cao
(c) Answer: (e³ - 2, 0)
(d) B1 correct increasing, concave-down logarithmic shape
(d) B1 vertical asymptote drawn at x = -2, with curve approaching but not crossing it
(d) B1ft both axis-crossing points correctly plotted, ft from parts (b) and (c)
(d) Answer: sketch showing asymptote x=-2, y-intercept (0,-2.31), x-intercept (e³-2, 0)

Question 9

(a) M1 splits log₂(4x³) as log₂(4) + log₂(x³)
(a) M1 uses log₂(4) = 2 and log₂(x³) = 3p
(a) A1 2 + 3p oe cao
(a) Answer: 2 + 3p
(b) M1 combines logs: log₃((x+5)/(x-1)) = 2
(b) M1 removes the log: (x+5)/(x-1) = 9
(b) M1 solves the resulting linear equation
(b) A1 x = 7/4 oe cao
(b) B1 states restriction x > 1 (required for both logs to be defined), consistent with x = 7/4
(b) Answer: x = 7/4, with restriction x > 1

Question 10

(a) M1 combines first equation: log₂(xy) = 5, so xy = 32
(a) M1 combines second equation: log₂(x/y) = 1, so x/y = 2, i.e. x = 2y
(a) M1 substitutes x = 2y into xy = 32 to form 2y² = 32
(a) A1 y = 4 cao
(a) A1 x = 8 cao
(a) B1 rejects y = -4 since log₂(y) requires y > 0
(a) Answer: x = 8, y = 4

Question 11

(a) B1 takes log₁₀ of both sides correctly: log₁₀(y) = log₁₀(A) + n*log₁₀(x)
(a) B1 identifies this as linear in the form Y = mX + c, with Y = log₁₀(y), X = log₁₀(x), gradient n and intercept log₁₀(A)
(a) Answer: log₁₀(y) = n*log₁₀(x) + log₁₀(A), which is linear with gradient n and intercept log₁₀(A)
(b) M1 calculates the gradient of the line using the two given points
(b) A1 n = 1.20 cao
(b) Answer: n = 1.20
(c) M1 identifies the y-intercept of the line as log₁₀(A) = 1.20
(c) M1 rearranges to A = 10^1.20
(c) A1 A = awrt 15.8
(c) Answer: A = awrt 15.8
(d) B1ft y = 15.8*x^1.2, ft from parts (b) and (c)
(d) Answer: y = 15.8x^1.2

Question 12

(a) M1 uses the substitution u = log₂(x) to form u² - 5u + 6 = 0
(a) M1 factorises as (u-2)(u-3) = 0
(a) A1 u = 2 and u = 3
(a) M1 converts back to x using log₂(x) = u
(a) A1 x = 4 (exact)
(a) A1 x = 8 (exact) cao
(a) Answer: x = 4 or x = 8

Question 13

(a) M1 states e^x = 3 + 2sqrt(2) directly from the definition of x
(a) M1 writes e^-x = 1/(3 + 2sqrt(2))
(a) M1 rationalises the denominator by multiplying by (3 - 2sqrt(2))/(3 - 2sqrt(2))
(a) A1 e^-x = 3 - 2sqrt(2) cao, since (3+2sqrt2)(3-2sqrt2) = 9 - 8 = 1
(a) A1 e^x + e^-x = (3+2sqrt2) + (3-2sqrt2) = 6, cso, answer printed
(a) Answer: 6 (shown)