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Calculate an approximation of log 7 in terms of log 5

by RoRi

With reference to the previous three exercises (here, here, and here) use x = \frac{1}{6} to calculate \log 7 in terms of \log 5. Obtain the qpproximation

    \[ 1.945907 < \log 7 < 1.945911. \]

Letting x = \frac{1}{6} we have

    \[ \log \frac{1+\frac{1}{6}}{1 - \frac{1}{6}} = \log \frac{7}{5} = \log 7 - \log 5. \]

Therefore, we apply Theorem 6.5,

    \begin{align*} &&\log \frac{7}{5} &= 2 \left( \frac{1}{6} + \frac{(1/6)^3}{3} + \frac{(1/6)^5}{5} + \frac{(1/6)^7}{7} \right) + R_4 \left( \frac{1}{6} \right) \\ \implies && \log 7 &= \log 5 + 2 \left( \frac{1}{6} + \frac{(1/6)^3}{3} + \frac{(1/6)^5}{5} + \frac{(1/6)^7}{7} \right) + R_4 \left( \frac{1}{6} \right) \\ \implies && 1.945907 &< \log 7 < 1.945911. \end{align*}

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