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How do exponential functions and logarithms describe growth and let us solve equations with the variable in the power?

Exponential functions and the number ee, the laws of logarithms and the relationship between exponentials and logarithms, solving equations of the form ax=ba^x = b, and using logarithms to linearise data.

A CCEA A-Level Mathematics answer on exponential functions and the number e, the laws of logarithms, the inverse relationship between exponentials and logarithms, solving equations of the form a to the x equals b, and using logarithms to linearise exponential and power data.

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What this dot point is asking

CCEA wants you to understand exponential functions and the special number ee, use the laws of logarithms, exploit the fact that logarithm and exponential are inverse operations to solve equations with the variable in the power, and use logarithms to linearise exponential or power-law data so a graph becomes a straight line. Exponential models of growth and decay run through both the pure and the applied content.

The answer

Exponential functions and the number e

The laws of logarithms

Solving equations of the form a^x = b

Because logarithm and exponential are inverse operations, taking logs of both sides of ax=ba^x = b brings the power down: xloga=logbx\log a = \log b, so x=logblogax = \dfrac{\log b}{\log a}. For equations involving ee, use the natural logarithm.

Linearising data with logarithms

Worked example: finding constants from a log graph

Examples in context

Example 1. Compound interest. An investment growing at 5%5\% a year follows A=P(1.05)tA = P(1.05)^t. To find when it doubles, set 2=1.05t2 = 1.05^t and take logs: t=log2/log1.0514.2t = \log 2 / \log 1.05 \approx 14.2 years. The logarithm is the only way to release a variable trapped in the exponent.

Example 2. Radioactive decay. A decaying mass m=m0eλtm = m_0 e^{-\lambda t} has a constant half-life found from 12=eλt1/2\tfrac{1}{2} = e^{-\lambda t_{1/2}}, giving t1/2=ln2/λt_{1/2} = \ln 2 / \lambda. The same exponential-then-log technique appears in physics, finance and biology.

Try this

Q1. Write log8+log5\log 8 + \log 5 as a single logarithm. [1 mark]

  • Cue. log40\log 40.

Q2. Solve 2x=502^x = 50, to three significant figures. [3 marks]

  • Cue. x=log50/log25.64x = \log 50 / \log 2 \approx 5.64.

Q3. Simplify ln(e3)\ln(e^3). [1 mark]

  • Cue. 33.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20214 marksSolve 32x+1=203^{2x + 1} = 20, giving your answer to three significant figures.
Show worked answer →

Take logarithms of both sides:

log(32x+1)=log20(2x+1)log3=log20.\log(3^{2x + 1}) = \log 20 \Rightarrow (2x + 1)\log 3 = \log 20.

So 2x+1=log20log3=1.30100.4771=2.7272x + 1 = \frac{\log 20}{\log 3} = \frac{1.3010}{0.4771} = 2.727.

Then 2x=1.7272x = 1.727, so x=0.864x = 0.864 (to three significant figures).

Markers reward taking logs, bringing the power down using the power law, isolating xx, and the correctly rounded answer.

CCEA 20195 marksThe mass mm grams of a sample decays so that m=50e0.2tm = 50e^{-0.2t}, where tt is the time in hours. Find the mass after 5 hours, and the time for the mass to fall to 10 grams.
Show worked answer →

After 55 hours: m=50e0.2×5=50e1=50×0.3679=18.4gm = 50e^{-0.2 \times 5} = 50e^{-1} = 50 \times 0.3679 = 18.4\,\text{g}.

For m=10m = 10: 10=50e0.2t10 = 50e^{-0.2t}, so e0.2t=0.2e^{-0.2t} = 0.2.

Taking natural logs: 0.2t=ln0.2=1.609-0.2t = \ln 0.2 = -1.609, so t=8.05hourst = 8.05\,\text{hours}.

Markers reward substituting t=5t = 5, the value 18.4g18.4\,\text{g}, rearranging to isolate the exponential, taking natural logs, and the time 8.058.05 hours.

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