Chapter 3.3, Problem 105E

To determine

To find:For which integers between 1 and 20 can you approximate natural logarithms, given the values $\ln 2\approx 0.6931$ , $\ln 3\approx 1.0986$ and $\ln 5\approx 1.6094$

Answer to Problem 105E

  $\begin{array}{l}\ln 1=0\\ \ln 2\approx 0.6931\\ \ln 3\approx 1.0986\\ \ln 4\approx 1.3862\end{array}$

  $\begin{array}{l}\ln 5\approx 1.6094\\ \ln 6\approx 1.7917\\ \ln 8\approx 2.0793\\ \ln 9\approx 2.1972\end{array}$

  $\begin{array}{l}\ln 10\approx 2.3025\\ \ln 12\approx 2.4848\\ \ln 15\approx 2.7080\end{array}$

  $\begin{array}{l}\ln 16\approx 2.7724\\ \ln 18\approx 2.8903\\ \ln 20\approx 2.9956\end{array}$

Explanation of Solution

Concept used:

Product property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a\left(uv\right)={\log }_{a}u+{\log }_{a}v$

Using natural logarithms of the integers 2, 3 and 5, the natural logarithms of only those integers between 1 and 20 can be determined whose prime factors contains only 2, 3 or 5.

These numbers are:1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, and 20.

Consider,

  $\ln 1=0$

Using the product property,

  $\begin{array}{c}\ln 4=\ln \left(2\times 2\right)\\ =\ln 2+\ln 2\\ \approx 0.6931+0.6931\\ \approx 1.3862\end{array}$

  $\begin{array}{c}\ln 6=\ln \left(2\times 3\right)\\ =\ln 2+\ln 3\\ \approx 1.7917\end{array}$

  $\begin{array}{c}\ln 8=\ln \left(2\times 4\right)\\ =\ln 2+\ln 4\\ \approx 2.0793\end{array}$

  $\begin{array}{c}\ln 9=\ln \left(3\times 3\right)\\ =\ln 3+\ln 3\\ \approx 2.1972\end{array}$

  $\begin{array}{c}\ln 10=\ln \left(2\times 5\right)\\ =\ln 2+\ln 5\\ \approx 2.3025\end{array}$

  $\begin{array}{c}\ln 12=\ln \left(2\times 6\right)\\ =\ln 2+\ln 6\\ \approx 2.4848\end{array}$

  $\begin{array}{c}\ln 15=\ln \left(3\times 5\right)\\ =\ln 3+\ln 5\\ \approx 2.7080\end{array}$

  $\begin{array}{c}\ln 16=\ln \left(2\times 8\right)\\ =\ln 2+\ln 8\\ \approx 2.7724\end{array}$

  $\begin{array}{c}\ln 18=\ln \left(2\times 9\right)\\ =\ln 2+\ln 9\\ \approx 2.8903\end{array}$

  $\begin{array}{c}\ln 20=\ln \left(2\times 10\right)\\ =\ln 2+\ln 10\\ \approx 2.9956\end{array}$