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IB Mathematics: Applications and Interpretation HL · 12.6: Exponents & Logarithms · 12.6.2

Logarithms Problem set A Worked solutions

Skills

  1. Solve:

    1. (a) \(\log_{10} x = -3\);

      • \(x = 10^{-3}\)
      • \(x = \frac{1}{1000}\)

    2. (b) \(\log_{10}(y-4) = -2\);

      • \(y - 4 = 10^{-2}\)
      • \(y = \frac{401}{100}\)

    3. (c) \(\log_{10}(2t+1) = 1\).

      • \(2t + 1 = 10^1\)
      • \(2t + 1 = 10 \Rightarrow 2t = 9\)
      • \(t = \frac{9}{2}\)

  2. Solve. Give the exact answer, then x to 3 significant figures.

    1. (a) \(\ln x = 3\);

      • \(x = e^3\)
      • From the calculator: \(x = 20.1\)

    2. (b) \(4\ln x + 10 = 0\);

      • \(4\ln x = -10 \Rightarrow \ln x = -2.5\)
      • \(x = e^{-2.5} = 0.0821\)

    3. (c) \(\ln(4x-3) = 2\).

      • \(4x - 3 = e^2\)
      • \(x = \frac{e^2+3}{4} = 2.60\)

  3. \(\text{Write each in terms of } p = \log_{10} a \text{ and } q = \log_{10} b:\)

    1. (a) \(\log_{10}\left(\frac{a^5}{b^2}\right)\);

      • \(\log a^5 - \log b^2\)
      • \(5 p - 2 q\)

    2. (b) \(\log_{10}\left(\sqrt[3]{a^2 b}\right)\);

      • \(\tfrac{1}{3}\left(2\log a + \log b\right)\)
      • \(\frac{2 p}{3} + \frac{q}{3}\)

    3. (c) \(\log_{10}\left(\frac{1000\sqrt{a}}{b^3}\right)\).

      • \(\log 1000 + \tfrac{1}{2}\log a - 3\log b\)
      • \(\frac{p}{2} - 3 q + 3\)

  4. Solve. Give exact answers.

    1. (a) \(\log_{10} x + \log_{10} 8 = 2\);

      • \(\log_{10}(8x) = 2\)
      • \(8x = 100\)
      • \(x = \frac{25}{2}\)

    2. (b) \(\log_{10}(x+4) + \log_{10} 5 = 2\).

      • \(\log_{10}\left(5(x+4)\right) = 2\)
      • \(5(x+4) = 100\)
      • \(x = 16\)

  5. Solve. Give exact answers.

    1. (a) \(\log_{10}(x+6) - \log_{10}(x-3) = 1\);

      • \(\frac{x+6}{x-3} = 10\)
      • \(x + 6 = 10x - 30 \Rightarrow 36 = 9x\)
      • \(x = 4\)

    2. (b) \(\ln(2x+1) - \ln(x-1) = 3\). Give x to 3 significant figures as well.

      • \(\frac{2x+1}{x-1} = e^3 \Rightarrow 2x + 1 = e^3x - e^3\)
      • \(e^3 + 1 = x(e^3 - 2)\)
      • \(x = \frac{e^3+1}{e^3-2} = 1.17\)

  6. Solve. Give the answer in terms of natural logarithms, then to 3 significant figures.

    1. (a) \(4^{x-1} = 3^{x+2}\);

      • \((x-1)\ln 4 = (x+2)\ln 3\)
      • \(x\ln 4 - x\ln 3 = 2\ln 3 + \ln 4\)
      • \(x = \frac{2\ln 3 + \ln 4}{\ln 4 - \ln 3} = 12.5\)

    2. (b) \(2^{3x+1} = 7^{x}\).

      • \((3x+1)\ln 2 = x\ln 7\)
      • \(3x\ln 2 + \ln 2 = x\ln 7 \Rightarrow \ln 2 = x(\ln 7 - 3\ln 2)\)
      • \(x = \frac{\ln 2}{\ln 7 - 3\ln 2} = -5.19\)

  7. Solve:

    1. (a) \(10^x = 0.04\). Give x to 3 significant figures.

      • \(x = \log_{10}0.04\)
      • From the calculator: \(x = −1.40\)

    2. (b) \(\log_{10}(3x+1) = 2\).

      • \(3x + 1 = 10^2 = 100\)
      • \(3x = 99, so x = 33\)

    3. (c) \(5\log_{10}x - 1 = 3\). Give x to 3 significant figures.

      • \(5\log_{10}x = 4\)
      • \(\log_{10}x = 0.8\)
      • \(x = 10^{0.8} = 6.31\)

  8. Solve:

    1. (a) \(e^x = 30\). Give x to 3 significant figures.

      • \(x = \ln 30\)
      • From the calculator: \(x = 3.40\)

    2. (b) \(2\ln x + 1 = 7\). Give the exact value of x, then x to 3 significant figures.

      • \(2\ln x = 6\)
      • \(\ln x = 3\)
      • \(x = e^3 = 20.1\)

    3. (c) \(5 - \ln x = 1.5\). Give x to 3 significant figures.

      • \(\ln x = 3.5\)
      • \(x = e^{3.5} = 33.1\)

In context

  1. The pH of a solution is \(\text{pH} = -\log_{10} h\), where h is the concentration of hydrogen ions in mol/dm³.

    1. (a) \(\text{Find the pH when } h = 3 \times 10^{-5}.\)

      • From the calculator: \(-\log_{10}(3 \times 10^{-5}) = 4.52\)

    2. (b) A different solution has pH 4.2. Find h.

      • \(\log_{10} h = -4.2\)
      • \(h = 10^{-4.2} = 6.31 \times 10^{-5}\)

  2. The height of a young tree is \(H = 2 + 1.5\ln t\) metres, where t is the age in years.

    1. (a) \(\text{Find the height when } t = 10.\)

      • From the calculator: \(2 + 1.5\ln 10 = 5.45\)

    2. (b) Find the age when the height is 5 m. Give an exact answer, then 3 significant figures.

      • \(1.5\ln t = 3 \Rightarrow \ln t = 2\)
      • \(t = e^2 = 7.39\)

  3. The population of town X is \(P = 12\,000 \times 1.03^t\) and the population of town Y is \(Q = 18\,000 \times 0.99^t\), where t is the number of years from now. Find when the two populations are equal. Give the exact form, then t to 3 significant figures.

    • \(12\,000 \times 1.03^t = 18\,000 \times 0.99^t\)
    • \(\frac{1.03^t}{0.99^t} = 1.5 \Rightarrow t\ln\!\left(\frac{1.03}{0.99}\right) = \ln 1.5\)
    • \(t = \frac{\ln 1.5}{\ln(1.03/0.99)} = 10.2\)

  4. The magnitude of an earthquake is \(M = \log_{10}A + 1.7\), where A is the largest ground movement in micrometres.

    1. (a) Find the magnitude when \(A = 4000\).

      • \(M = \log_{10}4000 + 1.7\)
      • From the calculator: \(M = 5.30\)

    2. (b) Find A when the magnitude is 6.2. Give your answer to 3 significant figures.

      • \(\log_{10}A + 1.7 = 6.2\), so \(\log_{10}A = 4.5\)
      • \(A = 10^{4.5} = 31 600\)

  5. The charge of a phone battery, \(C\%\), t hours after it is plugged in is modelled by \(C = 20 + 30\ln t\), for \(1 \le t \le 8\).

    1. (a) Find the charge after 3 hours.

      • \(C = 20 + 30\ln 3\)
      • From the calculator: \(C = 53.0\)

    2. (b) Find the exact time when the charge is 80%, and the time to 3 significant figures.

      • \(20 + 30\ln t = 80\), so \(\ln t = 2\)
      • \(t = e^2 = 7.39\)

  6. The depth D cm of sediment on a lake bed is modelled by \(D=a+b\ln t\), where t is the time in years since the lake was cleared. The depth is 4.0 cm when \(t=3\) and 6.5 cm when \(t=20\).

    1. (a) Find b and a. Give each to 3 significant figures.

      • \(a+b\ln3=4.0\) and \(a+b\ln20=6.5\)
      • Subtract: \(b\ln\dfrac{20}{3}=2.5\), so \(b=1.32\)
      • \(a=4.0-1.32\ln3=2.55\)

    2. (b) Find the depth predicted after 50 years.

      • \(D=2.55+1.32\ln50\)
      • From the calculator: \(D=7.71\) cm

    3. (c) Find when the depth is 8.0 cm.

      • \(2.55+1.32\ln t=8.0\), so \(\ln t=\dfrac{8.0-2.55}{1.32}\)
      • \(t=e^{(8.0-a)/b}=62.4\) years (use the unrounded a and b)

  7. The energy E joules released by an earthquake of magnitude M satisfies \(\log_{10}E=4.8+1.5M\).

    1. (a) Find the energy released by an earthquake of magnitude 6. Give your answer to 3 significant figures.

      • \(\log_{10}E=4.8+1.5\times6=13.8\)
      • \(E=10^{13.8}=6.31\times10^{13}\)

    2. (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 \(M=7.00\)

    3. (c) How many times more energy does a magnitude 8 earthquake release than a magnitude 6 earthquake?

      • \(\log_{10}E_8-\log_{10}E_6=1.5\times2=3\)
      • \(\dfrac{E_8}{E_6}=10^{3}=1000\)

Stretch

  1. Solve \(\log_{10} x + \log_{10}(x-3) = 1\).

    • \(\log_{10}\left(x(x-3)\right) = 1 \Rightarrow x^2 - 3x = 10\)
    • \(x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0\)
    • \(x = 5 \text{ or } x = -2. \text{ A log needs a positive number, so reject } x = -2.\)
    • \(x = 5\)

  2. Show that the solution of \(5^{x+1} = 3^{2x-1}\) can be written as \(x = \frac{\ln 15}{\ln 1.8}\). Then find x to 3 significant figures.

    • \((x+1)\ln 5 = (2x-1)\ln 3 \Rightarrow \ln 5 + \ln 3 = x(2\ln 3 - \ln 5)\)
    • \(x = \frac{\ln 5 + \ln 3}{2\ln 3 - \ln 5} = \frac{\ln 15}{\ln 9 - \ln 5} = \frac{\ln 15}{\ln 1.8}\)
    • From the calculator: \(x = 4.61\)

jmaths.xyz · AI HL · 12.6.2 Logarithms and solving exponential and log equations