Calculate an approximation of log 3 in terms of log 2

Use Theorem 6.5 (pp. 240-241 of Apostol) and the previous exercise to compute $\log 3$ in terms of $\log 2$ . To do this, observe that if $x = \frac{1}{5}$ then

$\frac{1+x}{1-x} = \frac{3}{2}.$

Using this value of $x$ and $m = 5$ establish the inequalities:

$1.098611 < \log 3 < 1.098617.$

Taking $x = \frac{1}{5}$ and $m = 5$ in Theorem 6.5 we have

$\begin{align*} \log \left( \frac{3}{2} \right) &= 2 \left( \frac{1}{5} + \frac{(1/5)^3}{3} + \frac{(1/5)^5}{5} + \frac{(1/5)^7}{7} + \frac{(1/5)^9}{9} \right) + R_5 \left( \frac{1}{5} \right) \\ &= 0.405465 + R_5 \left( \frac{1}{5} \right). \end{align*}$

Then, since $\log \left(\frac{3}{2} \right) = \log 3 - \log 2$ , we combine with the previous exercise (Section 6.11, Exercise #1) to obtain,

$1.098611 < \log 3 < 1.098617.$

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Calculate an approximation of log 3 in terms of log 2

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