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IB Mathematics: Applications and Interpretation HL · 12.6: Exponents & Logarithms · 12.6.2
Logarithms Problem set B One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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Solve:
(a) \(\log_{10} x = -4\);
- \(x = 10^{-4}\)
- \(x = \frac{1}{10000}\)
(b) \(\log_{10}(y-6) = -3\);
- \(y - 6 = 10^{-3}\)
- \(y = \frac{6001}{1000}\)
(c) \(\log_{10}(4t-3) = 1\).
- \(4t - 3 = 10^1\)
- \(4t - 3 = 10 \Rightarrow 4t = 16\)
- \(t = \frac{13}{4}\)
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Solve. Give the exact answer, then x to 3 significant figures.
(a) \(\ln x = -2\);
- \(x = e^{-2}\)
- From the calculator: \(x = 0.135\)
(b) \(6\ln x + 9 = 0\);
- \(6\ln x = -9 \Rightarrow \ln x = -1.5\)
- \(x = e^{-1.5} = 0.223\)
(c) \(\ln(5x+2) = 3\).
- \(5x + 2 = e^3\)
- \(x = \frac{e^3-2}{5} = 3.62\)
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\(\text{Write each in terms of } p = \log_{10} a \text{ and } q = \log_{10} b:\)
(a) \(\log_{10}\left(\frac{a^4}{b^5}\right)\);
- \(\log a^4 - \log b^5\)
- \(4 p - 5 q\)
(b) \(\log_{10}\left(\sqrt[4]{a b^3}\right)\);
- \(\tfrac{1}{4}\left(\log a + 3\log b\right)\)
- \(\frac{p}{4} + \frac{3 q}{4}\)
(c) \(\log_{10}\left(\frac{10\,000\sqrt[3]{b}}{a^2}\right)\).
- \(\log 10\,000 + \tfrac{1}{3}\log b - 2\log a\)
- \(- 2 p + \frac{q}{3} + 4\)
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Solve. Give exact answers.
(a) \(\log_{10} x + \log_{10} 4 = 3\);
- \(\log_{10}(4x) = 3\)
- \(4x = 1000\)
- \(x = 250\)
(b) \(\log_{10}(x+1) + \log_{10} 8 = 2\).
- \(\log_{10}\left(8(x+1)\right) = 2\)
- \(8(x+1) = 100\)
- \(x = \frac{23}{2}\)
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Solve. Give exact answers.
(a) \(\log_{10}(x+8) - \log_{10}(x-2) = 1\);
- \(\frac{x+8}{x-2} = 10\)
- \(x + 8 = 10x - 20 \Rightarrow 28 = 9x\)
- \(x = \frac{28}{9}\)
(b) \(\ln(3x+2) - \ln(x-1) = 2\). Give x to 3 significant figures as well.
- \(\frac{3x+2}{x-1} = e^2 \Rightarrow 3x + 2 = e^2x - e^2\)
- \(e^2 + 2 = x(e^2 - 3)\)
- \(x = \frac{e^2+2}{e^2-3} = 2.14\)
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Solve. Give the answer in terms of natural logarithms, then to 3 significant figures.
(a) \(5^{x-2} = 2^{x+3}\);
- \((x-2)\ln 5 = (x+3)\ln 2\)
- \(x\ln 5 - x\ln 2 = 3\ln 2 + 2\ln 5\)
- \(x = \frac{3\ln 2 + 2\ln 5}{\ln 5 - \ln 2} = 5.78\)
(b) \(3^{2x+1} = 11^{x}\).
- \((2x+1)\ln 3 = x\ln 11\)
- \(2x\ln 3 + \ln 3 = x\ln 11 \Rightarrow \ln 3 = x(\ln 11 - 2\ln 3)\)
- \(x = \frac{\ln 3}{\ln 11 - 2\ln 3} = 5.47\)
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Solve:
(a) \(10^x = 0.7\). Give x to 3 significant figures.
- \(x = \log_{10}0.7\)
- From the calculator: \(x = −0.155\)
(b) \(\log_{10}(2x-5) = 3\).
- \(2x - 5 = 10^3 = 1000\)
- \(2x = 1005, so x = \frac{1005}{2}\)
(c) \(4\log_{10}x + 2 = 5\). Give x to 3 significant figures.
- \(4\log_{10}x = 3\)
- \(\log_{10}x = 0.75\)
- \(x = 10^{0.75} = 5.62\)
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Solve:
(a) \(e^x = 12\). Give x to 3 significant figures.
- \(x = \ln 12\)
- From the calculator: \(x = 2.48\)
(b) \(3\ln x - 2 = 4\). Give the exact value of x, then x to 3 significant figures.
- \(3\ln x = 6\)
- \(\ln x = 2\)
- \(x = e^2 = 7.39\)
(c) \(8 - \ln x = 5.5\). Give x to 3 significant figures.
- \(\ln x = 2.5\)
- \(x = e^{2.5} = 12.2\)
In context
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The loudness of a sound is \(L = 10\log_{10}\!\left(\frac{I}{10^{-12}}\right)\) decibels, where I is the intensity in W/m².
(a) \(\text{Find L when } I = 3 \times 10^{-5}.\)
- From the calculator: \(10\log_{10}(3 \times 10^{7}) = 74.8\)
(b) A different sound has loudness 65 decibels. Find I.
- \(\log_{10}\!\left(\frac{I}{10^{-12}}\right) = 6.5\)
- \(I = 10^{-12} \times 10^{6.5} = 3.16 \times 10^{-6}\)
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The depth of mud in a lake bed is \(d = 0.4 + 2.5\ln t\) cm, where t is the time in years since the lake was dug.
(a) \(\text{Find the depth when } t = 20.\)
- From the calculator: \(0.4 + 2.5\ln 20 = 7.89\)
(b) Find the time when the depth is 6 cm. Give an exact answer, then 3 significant figures.
- \(2.5\ln t = 5.6 \Rightarrow \ln t = 2.24\)
- \(t = e^{2.24} = 9.39\)
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The number of bacteria in culture A is \(N = 500 \times 1.8^t\) and in culture B is \(M = 3200 \times 1.2^t\), where t is the time in hours. Find when the two numbers are equal. Give the exact form, then t to 3 significant figures.
- \(500 \times 1.8^t = 3200 \times 1.2^t\)
- \(\frac{1.8^t}{1.2^t} = 6.4 \Rightarrow t\ln 1.5 = \ln 6.4\)
- \(t = \frac{\ln 6.4}{\ln 1.5} = 4.58\)
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A camera's light meter shows \(E = \log_{10}L + 2.5\), where L is the light level in lux.
(a) Find the meter reading when \(L = 800\).
- \(E = \log_{10}800 + 2.5\)
- From the calculator: \(E = 5.40\)
(b) Find L when the reading is 7.1. Give your answer to 3 significant figures.
- \(\log_{10}L + 2.5 = 7.1\), so \(\log_{10}L = 4.6\)
- \(L = 10^{4.6} = 39 800\)
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The depth of snow, D cm, t hours after a storm begins is modelled by \(D = 15 + 12\ln t\), for \(t \ge 1\).
(a) Find the depth after 5 hours.
- \(D = 15 + 12\ln 5\)
- From the calculator: \(D = 34.3\)
(b) Find the exact time when the depth is 45 cm, and the time to 3 significant figures.
- \(15 + 12\ln t = 45\), so \(\ln t = 2.5\)
- \(t = e^{2.5} = 12.2\)
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The percentage P of a hillside covered in moss is modelled by \(P=a+b\ln t\), where t is the time in years since the hillside was cleared. The cover is 12% when \(t=2\) and 30% when \(t=15\).
(a) Find b and a. Give each to 3 significant figures.
- \(a+b\ln2=12\) and \(a+b\ln15=30\)
- Subtract: \(b\ln\dfrac{15}{2}=18\), so \(b=8.93\)
- \(a=12-8.93\ln2=5.81\)
(b) Find the percentage predicted after 40 years.
- \(P=5.81+8.93\ln40\)
- From the calculator: \(P=38.8\)%
(c) Find when the cover is 38%.
- \(5.81+8.93\ln t=38\), so \(\ln t=\dfrac{38-5.81}{8.93}\)
- \(t=e^{(38-a)/b}=36.7\) years (use the unrounded a and b)
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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) Find h when the pH is 3.4. Give your answer to 3 significant figures.
- \(\log_{10}h=-3.4\)
- \(h=10^{-3.4}=3.98\times10^{-4}\)
(b) Find the pH when \(h=2.5\times10^{-4}\).
- \(\text{pH}=-\log_{10}(2.5\times10^{-4})\)
- From the calculator: pH = 3.60
(c) How many times greater is the hydrogen ion concentration in a solution of pH 2 than in a solution of pH 5?
- \(\log_{10}h_{2}-\log_{10}h_{5}=-2-(-5)=3\)
- \(\dfrac{h_{2}}{h_{5}}=10^{3}=1000\)
Stretch
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Solve \(\log_{10} x + \log_{10}(x+3) = 1\).
- \(\log_{10}\left(x(x+3)\right) = 1 \Rightarrow x^2 + 3x = 10\)
- \(x^2 + 3x - 10 = 0 \Rightarrow (x+5)(x-2) = 0\)
- \(x = 2 \text{ or } x = -5. \text{ A log needs a positive number, so reject } x = -5.\)
- \(x = 2\)
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Show that the solution of \(7^{x+1} = 2^{3x-1}\) can be written as \(x = \frac{\ln 14}{\ln(8/7)}\). Then find x to 3 significant figures.
- \((x+1)\ln 7 = (3x-1)\ln 2 \Rightarrow \ln 7 + \ln 2 = x(3\ln 2 - \ln 7)\)
- \(x = \frac{\ln 7 + \ln 2}{3\ln 2 - \ln 7} = \frac{\ln 14}{\ln 8 - \ln 7} = \frac{\ln 14}{\ln(8/7)}\)
- From the calculator: \(x = 19.8\)
jmaths.xyz · AI HL · 12.6.2 Logarithms and solving exponential and log equations