Unit 1 Review Exercises

Exponents and Logarithms

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 Simplify \(\dfrac{\left(16x^{8}\right)^{\frac{3}{4}}}{(2x)^{3}}\).

Solution

\(\left(16x^{8}\right)^{\frac{3}{4}} = 16^{\frac{3}{4}}\,x^{6} = \left(\sqrt[4]{16}\right)^{3} x^{6} = 8x^{6}\), and \((2x)^{3} = 8x^{3}\).

\[ \frac{8x^{6}}{8x^{3}} = x^{3} \]

See 1.1.

2 Paper 1 Solve \(9^{x} = 27^{\,1 - x}\).

Solution

\(3^{2x} = 3^{3(1 - x)} \Rightarrow 2x = 3 - 3x \Rightarrow x = \dfrac{3}{5}\). See 1.1.

3 Paper 1 Solve \(2^{2x + 2} - 9\left(2^{x}\right) + 2 = 0\).

Solution

\(2^{2x + 2} = 4\left(2^{x}\right)^{2}\). Let \(u = 2^{x}\):

\[ 4u^{2} - 9u + 2 = 0 \;\Longrightarrow\; (4u - 1)(u - 2) = 0 \;\Longrightarrow\; u = \frac{1}{4} \;\text{ or }\; u = 2 . \]

\(2^{x} = 2^{-2} \Rightarrow x = -2\), and \(2^{x} = 2 \Rightarrow x = 1\). See 1.1.

4 Paper 1 Given that \(\log_5 2 = a\) and \(\log_5 3 = b\), express \(\log_5 7.2\) in terms of \(a\) and \(b\).

Solution

\(7.2 = \dfrac{36}{5} = \dfrac{2^{2} \times 3^{2}}{5}\), so

\[ \log_5 7.2 = 2\log_5 2 + 2\log_5 3 - \log_5 5 = 2a + 2b - 1 . \]

See 1.4.

5 Paper 1 Solve \(\log_4 x + \log_4 (x - 6) = 2\).

Solution

\(x(x - 6) = 4^{2} \Rightarrow x^{2} - 6x - 16 = 0 \Rightarrow (x - 8)(x + 2) = 0\).

\(x = -2\) makes \(\log_4 x\) undefined: reject. \(x = 8\). See 1.5.

6 Paper 1 Solve \(3^{x} = 5^{x - 2}\), giving your answer in the form \(\dfrac{\ln p}{\ln q}\), where \(p, q \in \mathbb{Q}\).

Solution

\(x\ln 3 = (x - 2)\ln 5 \Rightarrow 2\ln 5 = x(\ln 5 - \ln 3)\), so

\[ x = \frac{2\ln 5}{\ln 5 - \ln 3} = \frac{\ln 25}{\ln \frac{5}{3}} \;(\approx 6.30). \]

See 1.5.

7 Paper 2 A laptop costs $1800. Its value after \(t\) years is \(V = 1800 \times 0.78^{t}\).

(a) Write down the annual percentage decrease in value.

(b) Find the value after 3 years.

(c) Find the number of complete years after which the value is first less than $300.

Solution

(a) \(1 - 0.78 = 0.22\), so \(22\%\) per year.

(b) \(1800 \times 0.78^{3} = \$854.19\).

(c) Solve \(1800 \times 0.78^{t} = 300\): \(t = \dfrac{\ln \frac{1}{6}}{\ln 0.78} = 7.21\ldots\) The value is first less than $300 after 8 complete years.

See 1.2 and 1.5.

8 Paper 1 Let \(f(x) = 3 - \mathrm{e}^{-x}\).

(a) Find the \(y\)-intercept.

(b) Write down the horizontal asymptote. (c) Find the exact \(x\)-intercept.

(d) Write down the range of \(f\).

Solution

(a) \(f(0) = 3 - 1 = 2\), so \((0, 2)\). \(\qquad\) (b) \(y = 3\)

(c) \(\mathrm{e}^{-x} = 3 \Rightarrow -x = \ln 3 \Rightarrow x = -\ln 3\), so \((-\ln 3, 0)\).

(d) \(\mathrm{e}^{-x} > 0\), so \(f(x) < 3\). See 1.2.

9 Paper 2 The mass of the Earth is \(5.97 \times 10^{24}\) kg and the mass of the Moon is \(7.35 \times 10^{22}\) kg.

(a) How many times heavier is the Earth than the Moon? Give your answer to 3 s.f.

(b) Find the total mass of the Earth and the Moon, in standard form to 3 s.f.

Solution

(a) \(\dfrac{5.97 \times 10^{24}}{7.35 \times 10^{22}} = 81.2\) (3 s.f.).

(b) \(5.97 \times 10^{24} + 0.0735 \times 10^{24} = 6.0435 \times 10^{24} \approx 6.04 \times 10^{24}\) kg. See 1.3.

10 Paper 1 Given that \(\log_a x = 3\) and \(\log_a y = -2\), find

(a) \(\log_a \left(x^{2}\sqrt{y}\right)\) \(\qquad\) (b) \(\log_{a^{2}} (xy)\)

Solution

(a) \(2\log_a x + \dfrac{1}{2}\log_a y = 6 - 1 = 5\).

(b) Change of base to base \(a\):

\[ \log_{a^{2}} (xy) = \frac{\log_a (xy)}{\log_a a^{2}} = \frac{3 + (-2)}{2} = \frac{1}{2}. \]

See 1.4.