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IB Mathematics: Applications and Interpretation HL · 12.6: Exponents & Logarithms · 12.6.1
Laws of exponents Problem set A Worked solutions
Skills
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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.
- \(54x\)
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Find the exact value of:
(a) \(2^{-5}\).
- \(2^{-5} = \frac{1}{2^{5}}\)
- \(= \frac{1}{32}\)
(b) \(\left(\frac{3}{4}\right)^{-2}\).
- \(\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^{2}\)
- \(= \frac{16}{9}\)
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Find the exact value of:
(a) \(16^{\frac{3}{4}}\).
- \(16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^{3} = 2^{3}\)
- \(= 8\)
(b) \(32^{- \frac{2}{5}}\).
- \(\text{Negative index: take the reciprocal. } 32^{- \frac{2}{5}} = 1 \div 32^{\frac{2}{5}}\)
- \(32^{\frac{2}{5}} = \left(\sqrt[5]{32}\right)^{2} = 2^{2}\)
- \(= \frac{1}{4}\)
(c) \(\left(\frac{25}{36}\right)^{- \frac{3}{2}}\).
- \(\text{Negative index: flip the fraction. Use } \left(\frac{36}{25}\right)^{\frac{3}{2}}\)
- \(\left(\frac{36}{25}\right)^{\frac{3}{2}} = \left(\sqrt{\frac{36}{25}}\right)^{3} = \left(\frac{6}{5}\right)^{3}\)
- \(= \frac{216}{125}\)
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Write each expression in the form \(ax^p\):
(a) \(\frac{10x^4}{4\sqrt{x}}\);
- Write each root as a power of x, then use the index laws.
- \(\frac{10x^4}{4\sqrt{x}} = \frac{5}{2}x^{\frac{7}{2}}\)
(b) \(\frac{3\sqrt[3]{x}}{x^2}\);
- Write each root as a power of x, then use the index laws.
- \(\frac{3\sqrt[3]{x}}{x^2} = 3x^{- \frac{5}{3}}\)
(c) \(\sqrt{x} \times \sqrt[3]{x^2}\).
- Write each root as a power of x, then use the index laws.
- \(\sqrt{x} \times \sqrt[3]{x^2} = x^{\frac{7}{6}}\)
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Write each expression in the form \(ax^p + bx^q\):
(a) \(\frac{5}{\sqrt{x}} - 4x^2\sqrt[3]{x}\);
- Write each term as a power of x.
- \(-4x^{\frac{7}{3}} + 5x^{- \frac{1}{2}}\)
(b) \(\frac{9x^2-6\sqrt{x}}{3x^2}\).
- Divide each term on the top by the bottom, one at a time.
- \(3 - 2x^{- \frac{3}{2}}\)
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Solve:
(a) \(x^{\frac{3}{2}} = \frac{125}{27}\);
- \(x = \left(\frac{125}{27}\right)^{\frac{2}{3}}\)
- \(x = \frac{25}{9}\)
(b) \(x^{- \frac{2}{3}} = \frac{1}{4}\);
- \(x = \left(\frac{1}{4}\right)^{- \frac{3}{2}}\)
- \(x = 8\)
(c) \(\sqrt{x} = 4\sqrt[3]{x}\).
- \(\frac{x^{1/2}}{x^{1/3}} = 4 \Rightarrow x^{\frac{1}{6}} = 4\)
- \(x = 4^6 = 4096\)
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Solve:
(a) \(\left(\sqrt[3]{5}\right)^x = 25^{x-4}\);
- \(5^{\frac{x}{3}} = 5^{2(x-4)}\)
- \(\frac{x}{3} = 2x - 8\)
- \(x = \frac{24}{5}\)
(b) \(4^{x+1} = \left(\sqrt{8}\right)^{x+7}\).
- \(2^{2(x+1)} = 2^{\frac{3(x+7)}{2}}\)
- \(2(x+1) = \frac{3(x+7)}{2}\)
- \(x = 17\)
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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 \(4 - (-3) = 7\).
- \(\frac{5}{2}x^{7}\)
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Find the exact value of:
(a) \(4^{-3}\);
- \(4^{-3} = \frac{1}{4^3}\)
- \(= \frac{1}{64}\)
(b) \(\left(\frac{2}{5}\right)^{-3}\);
- \(\left(\frac{2}{5}\right)^{-3} = \left(\frac{5}{2}\right)^{3}\)
- \(= \frac{125}{8}\)
(c) \(5^0 + 3^{-1}\).
- \(5^0 = 1\) and \(3^{-1} = \frac{1}{3}\)
- \(1 + \frac{1}{3} = \frac{4}{3}\)
- \(= \frac{4}{3}\)
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Write each answer in the form \(a\times10^{k}\).
(a) Write 0.000 047 2 in standard form.
- Move the point 5 places right to make a = 4.72
- \(0.000\,047\,2=4.72\times10^{-5}\)
(b) Find \((4.5\times10^{6})\times(8\times10^{-9})\).
- \(4.5\times8=36\) and \(10^{6}\times10^{-9}=10^{-3}\)
- \(36\times10^{-3}=3.6\times10^{-2}\)
(c) Find \((7.2\times10^{8})+(5\times10^{9})\).
- \(7.2\times10^{8}=0.72\times10^{9}\)
- \(0.72\times10^{9}+5\times10^{9}=5.72\times10^{9}\)
(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
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The brightness of a lamp is \(B = 80d^{-2}\) lux at a distance of d metres. Find the exact brightness at a distance of:
(a) \(d = \frac{2}{5}\) m;
- \(80 \times \left(\frac{2}{5}\right)^{-2} = 80 \times \left(\frac{5}{2}\right)^{2}\)
- \(= 500\)
(b) \(d = 4\) m.
- \(80 \times 4^{-2} = \frac{80}{16}\)
- \(= 5\)
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A planet orbits the Sun at a distance of r astronomical units. Its orbital time is \(T = r^{3/2}\) years.
(a) Find T when r = 9.
- \(9^{3/2} = \left(\sqrt{9}\right)^3 = 3^3 = 27\)
(b) Find r when T = 125.
- \(r = 125^{2/3}\)
- \(= \left(\sqrt[3]{125}\right)^2 = 5^2 = 25\)
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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:
(a) 4 hours after noon;
- \(N = 3 \times 2^{4} = 3 \times 16\)
- \(= 48\)
(b) 3 hours before noon.
- \(N = 3 \times 2^{-3} = 3 \times \frac{1}{8}\)
- \(= \frac{3}{8}\)
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The mass of the Earth is \(5.97\times10^{24}\) kg and the mass of the Moon is \(7.35\times10^{22}\) kg.
(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
(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
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Answer both parts.
(a) Given that \(y = 125\sqrt{x^3}\), write \(\sqrt[3]{y}\) in the form \(ax^k\).
- \(y = 125x^{3/2}\)
- \(\sqrt[3]{y} = \left(125x^{\frac{3}{2}}\right)^{\frac{1}{3}}\)
- \(5x^{\frac{1}{2}}\)
(b) Given that \(y = x^2\sqrt[3]{x}\), write x in terms of y.
- \(y = x^{\frac{7}{3}}\)
- \(x = y^{1 \div \frac{7}{3}} = y^{\frac{3}{7}}\)
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Write \(\frac{\left(x + 2\sqrt{x}\right)^2}{x}\) in the form \(x^a + bx^c + d\).
- \(\left(x+2\sqrt{x}\right)^2 = x^2 + 4x\sqrt{x} + 4x\)
- Divide each term by x.
- \(x + 4x^{\frac{1}{2}} + 4\)
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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 HL · 12.6.1 Laws of exponents: negative and fractional indices