# Natural Logarithm

While logarithms can be in any base, most use cases involve either
the common logarithm in base 10 or the natural logarithm. The
natural logarithm has a base of **e**, the
mathematical constant approximately equal to 2.718 used frequently in science and
calculus. While the natural logarithm could theoretically be written
as
${\mathrm{log}}_{e}\left(x\right)$
, it's typically shortened to
${\mathrm{ln}}_{}\left(x\right)$
. We can convert any natural logarithm to
exponential form
using the following formula:

${\mathrm{ln}}_{}\left(x\right)=a\iff {e}^{a}=x$

In this article, we'll explore how to work with the natural logarithm both with and without a calculator. Let's get started!

## Working with the natural logarithm

The procedure for working with natural logarithms changes based on whether you're using a calculator. Most scientific calculators have dedicated ln functions, so solving problems is as easy as pressing the number inside the parenthesis and then "ln". For instance, if you wanted to find the ${\mathrm{ln}}_{}\left(7\right)$ , you would hit "7" followed by "ln" and receive an immediate answer of approximately 1.946.

If we don't have access to a scientific calculator, we'll have to convert the natural logarithm to exponential form and work it out. Returning to the example of ${\mathrm{ln}}_{}\left(7\right)$ , "e" is our base, 7 is the answer we're looking for, and "a" is the exponent we're raising "e" to. That gives us an equation of:

${e}^{a}=7$

${e}^{1.954}\approx 7$

Using the approximately equal sign is important because e is an irrational number. We cannot possibly calculate all of its digits, so most answers will only be approximations when working with natural logarithms. Some questions will ask us to round our answers to a specific decimal place.

Let's try working out a practice problem. Solve the equation and round your answer to the nearest thousandth:

${\mathrm{ln}}_{}\left(x\right)=-5.5$

First, we need to rewrite the equation in exponential form:

${e}^{-5.5}=x$

We'll need a calculator to work this out. Most calculators have a good approximation of e, but use 2.71828 if yours doesn't. The answer is:

$x\approx 0.004$

## The natural logarithm follows the properties of logarithms

The natural logarithm is a normal logarithm, meaning that all of the properties of logarithms we've learned apply to it. Let's use this to solve another practice problem:

Simplify: ${\mathrm{ln}}_{}\left({\left(3q\right)}^{2}\right)$

We can use the property to simplify logarithms of a power:

${\mathrm{log}}_{b}\left({x}^{y}\right)=y{\mathrm{log}}_{b}\left(x\right)$

This means that we can convert $\mathrm{ln}\left({\left(3q\right)}^{2}\right)=2\mathrm{ln}\left(3q\right)$ . Next, we can apply the property that the log of a product is equal to the sum of the logs:

${\mathrm{log}}_{b}\left(xy\right)={\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$

$2\mathrm{ln}\left(3q\right)=2\left(\mathrm{ln}\left(3\right)+\mathrm{ln}\left(q\right)\right)$

From here, we can plug the numbers into our calculator and solve the equation. The answer is approximately $2.197+2\mathrm{ln}\left(q\right)$ .

## Graphing the natural logarithm

The graph of $f\left(x\right)=\mathrm{ln}\left(x\right)$ looks similar to the graphs of other logarithmic functions such as $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ or $h\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ . The natural logarithm function has an asymptote at $x=0$ and an x-intercept at $\left(1,0\right)$ . It passes through the points $\left(\frac{1}{e},-1\right)$ and $\left(e,-1\right)$ as well. The diagram below highlights the natural logarithm function in blue while comparing it to several other logarithmic functions:

## Natural logarithm practice problems

a. Rewrite as a single logarithmic expression: $2\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+1\right)$

We need to use the properties of logarithms for this one, namely that $n\mathrm{ln}\left(a\right)=\mathrm{ln}\left({a}^{n}\right)$ and $\mathrm{ln}\left(a\right)-\mathrm{ln}\left(b\right)=\mathrm{ln}\left(\frac{a}{b}\right)$ . Using the first property, we get:

$2\mathrm{ln}\left(x\right)-\mathrm{ln}(x+1)=\mathrm{ln}\left({x}^{2}\right)-\mathrm{ln}(x+1)$

Using the second property, we get:

$\mathrm{ln}\left(\frac{{x}^{2}}{x+1}\right)$

We have successfully merged the two expressions.

b. What are the domain and range of $f\left(x\right)=\mathrm{ln}\left(x\right)$ ?

The natural logarithm is a logarithm of a positive number
(**e**), which means there is no power we can raise it to that
would yield a negative number. For instance,
${e}^{-3}$
represents
$\frac{1}{{e}^{-3}}$
, the ratio of two positive numbers will always be positive.
Furthermore, there is no power that we could raise **e** to that
would yield a zero. Therefore, the domain of the function excludes
zero and all negative numbers.

However, we can raise e to any power including zero and negative numbers. This means that the range of the function is all real numbers.

c. Solve for x and round your answer to the nearest tenth: $\mathrm{ln}(x+3)=3$

We need to give both sides of the equation the same base using e:

${e}^{\mathrm{ln}\left(x+3\right)}={e}^{3}$

Next, we need to recognize that the e and ln cancel each other out as inverse operations, leaving us with:

$x+3={e}^{3}$

From here, we simply solve for x:

$x={e}^{3}-3$

$x\approx 17.1$

## Topics related to the Natural Logarithm

## Flashcards covering the Natural Logarithm

## Practice tests covering the Natural Logarithm

College Algebra Diagnostic Tests

## Varsity Tutors can deepen a student's understanding of the natural logarithm

Outside of the common logarithm, the natural logarithm is the log that students see most often. Problems involving it usually don't say "log," so it's up to students to remember that they can apply all properties of logarithms to it. If you or your student need help with the natural logarithm or logarithms in general, an experienced tutor could help you identify how you study best and design lesson plans accordingly. Please contact the friendly Educational Directors at Varsity Tutors today to get started.

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