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IB Mathematics: Applications and Interpretation SL · 12.6: Exponents & Logarithms · 12.6.1

Laws of exponents Problem set A Worked solutions

Skills

  1. Write \((3x^2)^3 \times 2x^{-5}\) in the form \(ax^p\).

    • \((3x^2)^3 \times 2x^{-5} = 27x^6 \times 2x^{-5}\)
    • Multiply the numbers, add the powers of x.
    • \(54x\)
  2. Find the exact value of:

    1. (a) \(2^{-5}\).

      • \(2^{-5} = \frac{1}{2^{5}}\)
      • \(= \frac{1}{32}\)
    2. (b) \(\left(\frac{3}{4}\right)^{-2}\).

      • \(\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^{2}\)
      • \(= \frac{16}{9}\)
  3. Write \(\frac{(5x^2)^2}{10x^{-3}}\) in the form \(ax^p\).

    • \((5x^2)^2 = 25x^{4}\)
    • Divide the numbers and subtract the powers: \(25 \div 10\) and \(4 - (-3) = 7\).
    • \(\frac{5}{2}x^{7}\)
  4. Find the exact value of:

    1. (a) \(4^{-3}\);

      • \(4^{-3} = \frac{1}{4^3}\)
      • \(= \frac{1}{64}\)
    2. (b) \(\left(\frac{2}{5}\right)^{-3}\);

      • \(\left(\frac{2}{5}\right)^{-3} = \left(\frac{5}{2}\right)^{3}\)
      • \(= \frac{125}{8}\)
    3. (c) \(5^0 + 3^{-1}\).

      • \(5^0 = 1\) and \(3^{-1} = \frac{1}{3}\)
      • \(1 + \frac{1}{3} = \frac{4}{3}\)
      • \(= \frac{4}{3}\)
  5. Write each answer in the form \(a\times10^{k}\).

    1. (a) Write 0.000 047 2 in standard form.

      • Move the point 5 places right to make a = 4.72
      • \(0.000\,047\,2=4.72\times10^{-5}\)
    2. (b) Find \((4.5\times10^{6})\times(8\times10^{-9})\).

      • \(4.5\times8=36\) and \(10^{6}\times10^{-9}=10^{-3}\)
      • \(36\times10^{-3}=3.6\times10^{-2}\)
    3. (c) Find \((7.2\times10^{8})+(5\times10^{9})\).

      • \(7.2\times10^{8}=0.72\times10^{9}\)
      • \(0.72\times10^{9}+5\times10^{9}=5.72\times10^{9}\)
    4. (d) Find \(\dfrac{9\times10^{-3}}{1.5\times10^{-8}}\).

      • \(9\div1.5=6\) and \(10^{-3}\div10^{-8}=10^{5}\)
      • \(6\times10^{5}\)

In context

  1. The brightness of a lamp is \(B = 80d^{-2}\) lux at a distance of d metres. Find the exact brightness at a distance of:

    1. (a) \(d = \frac{2}{5}\) m;

      • \(80 \times \left(\frac{2}{5}\right)^{-2} = 80 \times \left(\frac{5}{2}\right)^{2}\)
      • \(= 500\)
    2. (b) \(d = 4\) m.

      • \(80 \times 4^{-2} = \frac{80}{16}\)
      • \(= 5\)
  2. The number of bacteria in a dish, in thousands, is \(N = 3 \times 2^t\), where t is the time in hours after noon. A negative t means before noon. Find the exact number of thousands of bacteria:

    1. (a) 4 hours after noon;

      • \(N = 3 \times 2^{4} = 3 \times 16\)
      • \(= 48\)
    2. (b) 3 hours before noon.

      • \(N = 3 \times 2^{-3} = 3 \times \frac{1}{8}\)
      • \(= \frac{3}{8}\)
  3. The mass of the Earth is \(5.97\times10^{24}\) kg and the mass of the Moon is \(7.35\times10^{22}\) kg.

    1. (a) Find their total mass, in the form \(a\times10^{k}\).

      • \(7.35\times10^{22}=0.0735\times10^{24}\)
      • \(5.97\times10^{24}+0.0735\times10^{24}=6.0435\times10^{24}\)
      • \(6.04\times10^{24}\) kg
    2. (b) How many times greater is the mass of the Earth than the mass of the Moon? Give your answer to 3 significant figures.

      • \(\dfrac{5.97\times10^{24}}{7.35\times10^{22}}\)
      • From the calculator: 81.2

Stretch

  1. Write \(\frac{(3x^{-2})^2 \times 8x^3}{6x^{-4}}\) in the form \(ax^p\).

    • \((3x^{-2})^2 = 9x^{-4}\)
    • Numerator: \(9 \times 8 = 72\) and \(x^{-4} \times x^3 = x^{-1}\).
    • Divide by \(6x^{-4}\): the number is \(72 \div 6\) and the power is \(-1 - (-4) = 3\).
    • \(12x^{3}\)

jmaths.xyz · AI SL · 12.6.1 Laws of exponents: negative and fractional indices