Homework for Thursday 1 October
Each part has its solution folded away. Do the exercise in your notebook and, once you have it, click on the statement to unfold the solution and check it.
The properties are in section 10.
Properties of logarithms
1
Find an approximation of the following logarithms
Use $\log 2 \approx 0.301$.
a) $\log 16$
$$\log 16 = \log 2^{4} = 4\log 2 \approx 4 \cdot 0.301 = 1.204$$
b) $\log \dfrac{1}{4}$
$$\log \frac{1}{4} = \log \frac{1}{2^{2}} = \log 2^{-2} = -2\log 2 \approx -2 \cdot 0.301 = -0.602$$
c) $\log \sqrt[3]{32}$
$$\log \sqrt[3]{32} = \log 2^{\frac{5}{3}} = \frac{5}{3}\log 2 \approx \frac{5}{3} \cdot 0.301 \approx 0.502$$
d) $\log \sqrt[4]{0.2}$
$0.2 = \frac{2}{10}$, and we know $\log 10 = 1$:
$$\log \sqrt[4]{\tfrac{2}{10}} = \log \left(\tfrac{2}{10}\right)^{\frac{1}{4}} = \frac{1}{4}\log \frac{2}{10} = \frac{1}{4}(\log 2 - \log 10)$$
$$\approx \frac{1}{4}(0.301 - 1) = \frac{-0.699}{4} \approx -0.175$$
e) $\log \sqrt[5]{\dfrac{1}{2}}$
$$\log \sqrt[5]{2^{-1}} = \log 2^{-\frac{1}{5}} = -\frac{1}{5}\log 2$$
$$\approx -\frac{1}{5} \cdot 0.301 \approx -0.0602$$
f) $\log 400$
$$\log 400 = \log(2^{2} \cdot 10^{2}) = \log 2^{2} + \log 10^{2} = 2\log 2 + 2\log 10$$
$$\approx 2 \cdot 0.301 + 2 \cdot 1 = 0.602 + 2 = 2.602$$
2
Write each expression as a single logarithm
a) $5\log_{2} A + 3\log_{2} B$
$$\log_{2} A^{5} + \log_{2} B^{3} = \log_{2}\left(A^{5} B^{3}\right)$$
b) $\dfrac{1}{2}\ln A + 2\ln B - 3\ln C$
$$\ln A^{\frac{1}{2}} + \ln B^{2} - \ln C^{3} = \ln \frac{A^{\frac{1}{2}} B^{2}}{C^{3}} = \ln \frac{\sqrt{A}\, B^{2}}{C^{3}}$$
c) $\dfrac{1}{2}\log A - \dfrac{3}{2}\log B$
$$\log A^{\frac{1}{2}} - \log B^{\frac{3}{2}} = \log \frac{A^{\frac{1}{2}}}{B^{\frac{3}{2}}} = \log \sqrt{\frac{A}{B^{3}}}$$
3
Find the unknown
Find $x$ using the properties of logarithms.
a) $\log_{5} x + \log_{5} 3 = \log_{5} 15$
$$\log_{5} 3x = \log_{5} 15 \;\Longrightarrow\; 3x = 15 \;\Longrightarrow\; x = 5$$
If two logarithms with the same base are equal, their arguments are equal too.
b) $\log x + 2\log 2 = 2\log 3 + \log 12$
$$\log x + \log 2^{2} = \log 3^{2} + \log 12$$
$$\log 4x = \log 108 \;\Longrightarrow\; 4x = 108 \;\Longrightarrow\; x = \frac{108}{4} = 27$$