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IB Mathematics: Applications and Interpretation SL · 12.6: Exponents & Logarithms · 12.6.2
Logarithms Problem set A Worked solutions
Skills
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Solve:
(a) \(\log_{10} x = -3\);
- \(x = 10^{-3}\)
- \(x = \frac{1}{1000}\)
(b) \(\log_{10}(y-4) = -2\);
- \(y - 4 = 10^{-2}\)
- \(y = \frac{401}{100}\)
(c) \(\log_{10}(2t+1) = 1\).
- \(2t + 1 = 10^1\)
- \(2t + 1 = 10 \Rightarrow 2t = 9\)
- \(t = \frac{9}{2}\)
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Solve. Give the exact answer, then x to 3 significant figures.
(a) \(\ln x = 3\);
- \(x = e^3\)
- From the calculator: \(x = 20.1\)
(b) \(4\ln x + 10 = 0\);
- \(4\ln x = -10 \Rightarrow \ln x = -2.5\)
- \(x = e^{-2.5} = 0.0821\)
(c) \(\ln(4x-3) = 2\).
- \(4x - 3 = e^2\)
- \(x = \frac{e^2+3}{4} = 2.60\)
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Solve:
(a) \(10^x = 0.04\). Give x to 3 significant figures.
- \(x = \log_{10}0.04\)
- From the calculator: \(x = −1.40\)
(b) \(\log_{10}(3x+1) = 2\).
- \(3x + 1 = 10^2 = 100\)
- \(3x = 99, so x = 33\)
(c) \(5\log_{10}x - 1 = 3\). Give x to 3 significant figures.
- \(5\log_{10}x = 4\)
- \(\log_{10}x = 0.8\)
- \(x = 10^{0.8} = 6.31\)
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Solve:
(a) \(e^x = 30\). Give x to 3 significant figures.
- \(x = \ln 30\)
- From the calculator: \(x = 3.40\)
(b) \(2\ln x + 1 = 7\). Give the exact value of x, then x to 3 significant figures.
- \(2\ln x = 6\)
- \(\ln x = 3\)
- \(x = e^3 = 20.1\)
(c) \(5 - \ln x = 1.5\). Give x to 3 significant figures.
- \(\ln x = 3.5\)
- \(x = e^{3.5} = 33.1\)
In context
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The pH of a solution is \(\text{pH} = -\log_{10} h\), where h is the concentration of hydrogen ions in mol/dm³.
(a) \(\text{Find the pH when } h = 3 \times 10^{-5}.\)
- From the calculator: \(-\log_{10}(3 \times 10^{-5}) = 4.52\)
(b) A different solution has pH 4.2. Find h.
- \(\log_{10} h = -4.2\)
- \(h = 10^{-4.2} = 6.31 \times 10^{-5}\)
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The height of a young tree is \(H = 2 + 1.5\ln t\) metres, where t is the age in years.
(a) \(\text{Find the height when } t = 10.\)
- From the calculator: \(2 + 1.5\ln 10 = 5.45\)
(b) Find the age when the height is 5 m. Give an exact answer, then 3 significant figures.
- \(1.5\ln t = 3 \Rightarrow \ln t = 2\)
- \(t = e^2 = 7.39\)
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The magnitude of an earthquake is \(M = \log_{10}A + 1.7\), where A is the largest ground movement in micrometres.
(a) Find the magnitude when \(A = 4000\).
- \(M = \log_{10}4000 + 1.7\)
- From the calculator: \(M = 5.30\)
(b) Find A when the magnitude is 6.2. Give your answer to 3 significant figures.
- \(\log_{10}A + 1.7 = 6.2\), so \(\log_{10}A = 4.5\)
- \(A = 10^{4.5} = 31 600\)
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The charge of a phone battery, \(C\%\), t hours after it is plugged in is modelled by \(C = 20 + 30\ln t\), for \(1 \le t \le 8\).
(a) Find the charge after 3 hours.
- \(C = 20 + 30\ln 3\)
- From the calculator: \(C = 53.0\)
(b) Find the exact time when the charge is 80%, and the time to 3 significant figures.
- \(20 + 30\ln t = 80\), so \(\ln t = 2\)
- \(t = e^2 = 7.39\)
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The energy E joules released by an earthquake of magnitude M satisfies \(\log_{10}E=4.8+1.5M\).
(a) Find the energy released by an earthquake of magnitude 6. Give your answer to 3 significant figures.
- \(\log_{10}E=4.8+1.5\times6=13.8\)
- \(E=10^{13.8}=6.31\times10^{13}\)
(b) Find the magnitude of an earthquake that releases \(2\times10^{15}\) joules.
- \(\log_{10}(2\times10^{15})=15.30\)
- \(1.5M=15.30-4.8\), so \(M=7.00\)
(c) How many times more energy does a magnitude 8 earthquake release than a magnitude 6 earthquake?
- \(\log_{10}E_8-\log_{10}E_6=1.5\times2=3\)
- \(\dfrac{E_8}{E_6}=10^{3}=1000\)
jmaths.xyz · AI SL · 12.6.2 Logarithms and solving exponential and log equations