Natural Logarithms

After $\pi$, the next weird number is called $e$, for $\textit{exponential}$. Jacob Bernoulli first discussed it in 1683. It occurs in problems about compound interest, leads to logarithms, and tells us how variables like radioactivity, temperature, or the human population increase or decrease.

In 1614 John Napier knew, from personal experience, that many scientific problems, especially in astronomy, required multiplying complicated numbers or finding square and cube roots. At a time when there was no electricity, let alone computers, calculations had to be done by hand. Adding two decimal numbers together was reasonably simple, but multiplying them was much harder. So Napier invented a method for turning multiplication into addition. The trick was to work with the powers of a fixed number.

Since many exponential models have base $e$, it is useful also to consider the logarithm in base $e$, called the $\textit{natural logarithm}$.

The $\textit{natural logarithm}$ of a positive number is its power of $e$.
The natural logarithm of $x$ is written $\log_{e}{x}$ or $\ln{x}$.

For example, $e^{2.1} \approx8.166 \cdots$, so $\log_{e}{8.166} \approx 2.1$, or $\ln{8.166} \approx 2.1$.
$$\large e^x = y \Leftrightarrow x = \log_{e}{y} \large$$

Example 1

Write an equivalent statement for $e^2 \approx 7.389$.

$\log_{e}{7.389} \approx 2$

Example 2

Write an equivalent statement for $\log_{e}{25} \approx 3.22$.

$e^{3.22} \approx 25$

From the rule $e^x = y \Leftrightarrow x = \log_{e}{y}$, we obtain that:
\begin{align} \require{AMSsymbols} \displaystyle e^x &= y \cdots (1) \\ x &= \ln{y} \cdots (2) \\ \color{green} x &\color{green}= \color{green}\ln{e^x} &\color{red}{\text{substitute } (1) \text{ into } (2)} \\ \color{green}e^{\ln{y}} &\color{green}= \color{green}y &\color{red}{\text{substitute } (2) \text{ into } (1)} \end{align}

Example 3

Simplify $\ln{e^4}$.

$\ln{e^4} = 4$

Example 4

Simplify $e^{\ln{5}}$.

$e^{\ln{5}} = 5$

Example 5

Simplify $\ln{\dfrac{1}{e^3}}$.

\begin{align} \displaystyle \ln{\dfrac{1}{e^3}} &= \ln{e^{-3}} \\ &= -3 \end{align}

Example 6

Find $x$, if $10=e^x$.

\begin{align} \displaystyle 10 &= e^{\ln{10}} \\ &= e^{2.30 \cdots} \\ \therefore x &= 2.30 \cdots \end{align}

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