Mathematics · OCR H240 · Topic 1.06

Exponentials and logarithms

Exponential functions such as eˣ, logarithms and their laws, and modelling growth and decay.

  • 7 sub-topics
  • Taught in Year 12
  • 101 practice questions
  • 7 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 1.06aProperties of the exponential functionY12
  2. 1.06bGradient of e^(kx)Y12
  3. 1.06c-eProperties of the logarithmY12
  4. 1.06fLaws of logarithmsY12
  5. 1.06gEquations involving exponentialsY12
  6. 1.06hReduction to linear formY12
  7. 1.06iModelling using exponential functionsY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 101 questions in this topic, free, no account needed.

Exponentials and logarithms

Easy

An exponential equation · Year 12 · 2 marks

  1. Solve 3x=203^x = 20, giving xx to 3 significant figures.
    (2)

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  1. Take logs of both sides: xln⁡3=ln⁡20x\ln 3 = \ln 20.x=ln⁡20ln⁡3x = \dfrac{\ln 20}{\ln 3}
  2. x=2.73x = 2.73

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Exponentials and logarithms

Medium

A hidden quadratic · Year 12 · 4 marks

  1. Solve 22x−5×2x+4=02^{2x} - 5 \times 2^x + 4 = 0.
    (4)
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  1. 22x=(2x)22^{2x} = (2^x)^2.Let y=2xy = 2^x: y2−5y+4=0y^2 - 5y + 4 = 0
  2. (y−1)(y−4)=0(y - 1)(y - 4) = 0
  3. 2x=1⇒x=02^x = 1 \Rightarrow x = 0; 2x=4⇒x=22^x = 4 \Rightarrow x = 2

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Exponentials and logarithms

Hard

Modelling cooling · Year 12 · 7 marks

A cup of tea is left in a room. Its temperature TT °C, tt minutes later, is modelled by T=20+70e−kt,T = 20 + 70\mathrm{e}^{-kt}, where kk is a positive constant. After 10 minutes the temperature is 50 °C.
  1. (a)
    Find the value of kk, to 3 significant figures.
    (3)
  2. (b)
    Find how long it takes for the tea to cool to 30 °C.
    (3)
  3. (c)
    Give one limitation of the model.
    (1)

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Part (a)3 marks

  1. Substitute t=10t = 10 and T=50T = 50, then isolate the exponential.5050=20+70e−10k{}= 20 + 70\mathrm{e}^{-10k}⇒e−10k{}\Rightarrow \mathrm{e}^{-10k}=37{}= \tfrac{3}{7}
  2. Take natural logs of both sides to bring down the power.−10k-10k=ln⁡37{}= \ln\tfrac{3}{7}⇒k{}\Rightarrow k=110ln⁡73{}= \tfrac{1}{10}\ln\tfrac{7}{3}
  3. Evaluate and round.k=0.0847k = 0.0847

Part (b)3 marks

  1. Substitute T=30T = 30 and isolate the exponential.e−kt=17\mathrm{e}^{-kt} = \tfrac{1}{7}
  2. Take logs; ln⁡17=−ln⁡7\ln\tfrac{1}{7} = -\ln 7.t=ln⁡7kt = \dfrac{\ln 7}{k}
  3. Use the unrounded value of kk to avoid rounding errors.t=23.0t = 23.0 minutes

Part (c)1 mark

  1. As tt grows, TT approaches 20, so 20 °C must be the room temperature.It assumes the room stays at exactly 20 °C.

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