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Calculate an approximation of log 5 in terms of log 2

by RoRi

With reference to the previous two exercises (here and here) obtain the following inequalities for \log 5,

    \[ 1.609435 < \log 5 < 1.609438. \]

If we take x = \frac{1}{9} then we have

    \[ \log \left( \frac{1+x}{1-x} \right) = \log \left( \frac{1+\frac{1}{9}}{1-\frac{1}{9}} \right) = \log 5 - 2 \log 2. \]

Therefore, we apply Theorem 6.5 to obtain

    \begin{align*} && &\log \frac{5}{4} = 2 \left( \frac{1}{9} + \frac{(1/9)^3}{3} + \frac{(1/9)^5}{5} + \frac{(1/9)^7}{7} \right) + R_4 \left( \frac{1}{9} \right) \\ \implies && &1.609435 < \log 5 < 1.609438. \end{align*}

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