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IB Mathematics: Applications and Interpretation SL · 12.6: Exponents & Logarithms · 12.6.3

Mixed exponent and logarithm practice 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

  1. Solve each equation.

    1. (a) \(\log_{10}(x-7)=3\)

      • \(x-7=10^3=1000\)
      • x = 1007
    2. (b) \(\log_{10}(4x+1)=2\)

      • \(4x+1=10^2=100\)
      • 4x = 99, so x = 24.75
    3. (c) \(\ln(x+2)=1.5\)

      • \(x+2=e^{1.5}\)
      • x = e^1.5 − 2 = 2.48
  2. Solve each equation by writing both sides as powers of the same base.

    1. (a) \(8^{x+1}=16^{2x-3}\)

      • \(8^{x+1}=2^{3x+3}\) and \(16^{2x-3}=2^{8x-12}\)
      • 3x + 3 = 8x − 12
      • x = 3
    2. (b) \(25^{2x-1}=125^{x+2}\)

      • \(25^{2x-1}=5^{4x-2}\) and \(125^{x+2}=5^{3x+6}\)
      • 4x − 2 = 3x + 6
      • x = 8
    3. (c) \(2^{3x+1}=\dfrac1{64}\)

      • \(\tfrac1{64}=2^{-6}\)
      • 3x + 1 = −6
      • \(x=-\tfrac73\)
  3. Solve each equation.

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

      • \(8x = 10^2 = 100\)
      • x = 12.5
    2. (b) \(\log_{10}(x-3) = -2\)

      • \(x - 3 = 10^{-2} = 0.01\)
      • x = 3.01
    3. (c) \(\ln(4x+3) = 2\)

      • \(4x + 3 = e^{2}\)
      • \(x = \frac{e^{2} - 3}{4} = 1.10\)
  4. Solve each equation by writing both sides as powers of the same base.

    1. (a) \(7^{x-1} = 343\)

      • \(343 = 7^3\), so \(x - 1 = 3\)
      • x = 4
    2. (b) \(4^{x} = 32\)

      • \(4 = 2^2\) and \(32 = 2^5\), so \(2^{2x} = 2^5\)
      • \(2x = 5\), so \(x = \frac{5}{2}\)
    3. (c) \(3^{2x-1} = 81^{x-3}\)

      • \(81 = 3^4\), so \(81^{x-3} = 3^{4(x-3)}\)
      • \(2x - 1 = 4x - 12\)
      • \(2x = 11\), so \(x = \frac{11}{2}\)

In context

  1. The value of a car is modelled by \(V=24\,000\times0.88^{t}\) euros, where t is the time in years after it is bought.

    1. (a) Find the value when it is bought.

      • t = 0: \(V=24\,000\times0.88^{0}=24\,000\)
    2. (b) Find the value after 3 years.

      • \(V=24\,000\times0.88^{3}\)
      • From the calculator: 16 355.33 euros
    3. (c) Use your calculator to find when the value has fallen to 10 000 euros.

      • \(24\,000\times0.88^{t}=10\,000\)
      • Solve for t: t = 6.85 years
  2. The number of members of an online forum is modelled by \(N=320\times1.35^{t}\), where t is the time in weeks.

    1. (a) Write down the number of members at the start, and find the number after 6 weeks. Give your answer to 3 significant figures.

      • t = 0: \(N=320\times1.35^{0}=320\)
      • \(N=320\times1.35^{6}\)
      • From the calculator: 1940
    2. (b) Use your calculator to find when there are 5000 members. Give t to 3 significant figures.

      • \(320\times1.35^{t}=5000\)
      • Solve for t: t = 9.16 weeks
    3. (c) Write down the percentage increase each week.

      • Each week the number is multiplied by 1.35.
      • 1.35 means an increase of 35%.

Stretch

  1. Solve \(8^{x} = \dfrac{4^{x+3}}{2}\).

    • \(8 = 2^3\), \(4 = 2^2\) and \(2 = 2^1\)
    • \(2^{3x} = \dfrac{2^{2(x+3)}}{2^1} = 2^{2x+5}\)
    • \(3x = 2x + 5\), so x = 5

jmaths.xyz · AI SL · 12.6.3 Mixed exponent and logarithm practice