# Calculus I, II, II, and differential equations College Park Tutors Blog

## An Overview of the Natural Logarithm: Common Questions and Mistakes Explained

Calculus 1, Math140
May 18, 2017

## What is the natural logarithm?

This part is optional, as students are unlikely to be tested on this material, but if you would like a better understanding of what the natural logarithm means conceptually, you may want to read this section.

Sometimes students will see $ln(x)$ on a paper, refer to it as "el-en", but not know what it actually means. Perhaps the easiest way to understand it is to know its relation to the number, $e$. Remember: $e$ is just a constant, approximately equal to $2.71828$. $e^x$ and $ln(x)$ are inverses.

What does this mean? Think about cubing a number and taking the cube root of a number. Imagine that we start out with the number $7$. If we cube it, we get $343$. If we take the cube root of this new number, $343$, we get $7$, which is the number we started with. Similarly, imagine that we start out with the number $125$. If we take the cube root, we get $5$. If we cube this new number, $5$, we get $125$, which is the number we started with. Cubing and taking the cube root are inverses-- they undo each other. If one of the operations occurs, in order to undo it, we simply apply the other. Stated more mathematically,

$x \Rightarrow \text{cube it} \Rightarrow x^3 \Rightarrow \text{take cube root} \Rightarrow \sqrt[3]{x^3} \Rightarrow \text{simplifies to x, what we started with}$

OR

$x \Rightarrow \text{take cube root} \Rightarrow \sqrt[3]{x} \Rightarrow \text{cube it} \Rightarrow ( \sqrt[3]{x} )^3 \Rightarrow \text{simplies to x, what we started with}$

The exponential function and the natural logarithmic function work similarly. So how does this help with solving problems? We can use the fact that these functions are inverses to solve for $x$. Take, for example, the problem $e^x = 5$. You can solve for $x$ by taking the natural log of both sides of the equation.

$e^x = 5 \\[2ex] ln(e^x) = ln(5) \\[2ex] x = ln(5)$

The key here is that $ln(e^x)$ simplifies to $x$, much like how taking the cube root of $x^3$ is simply $x$.

## What are natural logarithm properties?

$ln(ab) = ln(a) + ln(b) \\[2ex] ln\left(\frac{a}{b}\right) = ln(a) - ln(b) \\[2ex] ln(a^b) = bln(a)$

Here's something important to understand about the third natural log property: in order for it to apply, the exponent has to be within the natural logarithm. In other words, it would be incorrect to say, for example, that $ln(x)^2$ simplifies to $2ln(x)$. It would be correct, however, to say that $ln(x^2)$ simplifes to $2ln(x)$. Remember: $ln(x)^2$ and $ln(x^2)$ are different. The first one can be seen as $ln(x) \times ln(x)$, whereas the second one can be seen as $ln(x \times x)$, which is, by the first natural log property, $ln(x)+ln(x)$, which can be rewritten as $2ln(x)$. Clearly, these are two different expressions.

Also important to note is that you cannot simplify expressions of the forms $ln(a+b)$ nor $ln(a-b)$. Make sure you pay attention to that kind of detail, because it may be tempting to look at an expression in one of those forms and apply the first or second property, but doing so would be incorrect.

## How do you take derivatives with the natural logarithm?

You may be asked to take the derivative of a function like this:

$f(x) = ln\left(\frac{3x^4e^x}{\sqrt[]{5x-2}}\right)$

Applying the natural logarithm properties one at a time, we get:

$f(x) = ln\left(\frac{3x^4e^x}{\sqrt[]{5x-2}}\right) \\[2ex] = ln\left(3x^4 \times \frac{e^x}{\sqrt[]{5x-2}}\right) \\[2ex] = ln(3x^4) + ln\left(\frac{e^x}{\sqrt[]{5x-2}}\right) \\[2ex] = ln(3x^4) + ln(e^x) - ln(\sqrt[]{5x-2}) \\[2ex] = ln(3x^4) + x - ln(\sqrt[]{5x-2}) \\[2ex] = ln(3x^4) + x - ln((5x-2)^\frac{1}{2}) \\[2ex] = ln(3x^4) + x - \frac{1}{2}ln(5x-2)$

Now we are ready to take the derivative. Here's the rule:

$\text{If } f(x) = ln(x)\text{, then} f'(x)= \frac{1}{x}$

Now, you may encounter something more complicated, such as $f(x) = ln(x^5 +x^2)$. The key here is to understand that the chain rule applies. So the derivative would be $f'(x) = \frac{1}{x^5 + x^2}$ times the derivative of $x^5 + x^2$, which is $5x^4 + 2x$. So, you would have $\frac{1}{x^5 + x^2} \times (5x^4 + 2x) = \frac{5x^4 + 2x}{x^5 + x^2}$. Written more mathematically, if $f(x) = ln(g(x))$, then $f'(x) = \frac{1}{g(x)} \times g'(x)$, which can be rewritten as $\frac{g'(x)}{g(x)}$. Now, let's go back to the original problem.

$f(x)=ln(3x^4) + x - \frac{1}{2}ln(5x-2) \\[2ex] f'(x)= \frac{12x^3}{3x^4} + 1 - \frac{1}{2} \times \frac{5}{5x-2} \\[2ex] = \frac{4}{x} + 1 - \frac{5}{10x-4}$

And that's the answer.

## How do you do logarithmic differentiation?

Sometimes you may find that it is easier to take the derivative of a function by using natural logarithms instead of using approaches like the product and quotient rules, which can get messy pretty quickly.

Here's the process: You take the natural log of both sides. Then, you simplify as much as you'd like by using the natural log properties. Then, you use implicit differentiation (that's where $\frac{1}{y} \times \frac{dy}{dx}$ comes from in the example below). Then, you multiply both sides by the original function.

$y= \frac{e^x 3x^3}{ln(x)\sqrt[]{2x+7}} \\[2ex] ln(y)=ln\left(\frac{e^x 3x^3}{ln(x)(2x+7)^\frac{1}{2}}\right) \\[2ex] ln(y)=ln(e^x 3x^3) - ln(ln(x)(2x+7)^\frac{1}{2}) \\[2ex] ln(y)=ln(e^x)+ln(3x^3)-[ln(ln(x)) + ln((2x+7)^\frac{1}{2})] \\[2ex] ln(y)=x+ln(3x^3)-ln(ln(x)) - \frac{1}{2}ln(2x+7) \\[2ex] \frac{1}{y} \times \frac{dy}{dx} =1+ \frac{9x^2}{3x^3}-\frac{\frac{1}{x}}{ln(x)}-\frac{1}{2} \times \frac{2}{2x+7} \\[2ex] \frac{dy}{dx} = (1+\frac{9x^2}{3x^3} - \frac{\frac{1}{x}}{ln(x)} - \frac{1}{2} \times \frac{2}{2x+7}) \times y \\[2ex] \frac{dy}{dx}= (1+ \frac{9x^2}{3x^3} - \frac{\frac{1}{x}}{ln(x)} - \frac{1}{2} \times \frac{2}{2x+7}) \times \frac{e^x3x^3}{ln(x)\sqrt[]{2x+7}}$

Now, in the example above, logarithmic differentiation was optional. In theory, you could have used quotient rule, product rule, etc., but that would have been a pretty messy process. Think about it; you'd have to use product rule within quotient rule. That is messy, and you're more susceptible to making errors by using that method.

Sometimes, however, you don't even have a choice; you have to use logarithmic differentiation. Take a look at the following function.

$y = x^x$

Though it may be tempting, you can't use the power rule on this because the power rule applies only when the exponent is a constant. In this case, the exponent is a variable. So let's go ahead with the logarithmic differentiation.

## How do you solve integrals using the natural logarithm?

The power rule for integrals is probably the first integration rule you learned. The rule is that, if $f(x) = x^n$ where $n$ is some constant, then an antiderivative (let's call it $g(x)$) is $g(x) = \frac{x^{n+1}}{n+1}$. If you find that definition confusing, you can think of it this way: add $1$ to the exponent, and then divide the whole expression by that number. For example, an antiderivative of $x^7$ is $\frac{x^8}{8}$ because you add $1$ to $7$ (which makes $8$), and then you divide this new expression by that number. Even if the original function has a coefficient other than $1$, the rule still applies. For example, an antiderivative of $9x^5$ is $\frac{9x^6}{6}$, which is $\frac{3}{2}x^6$. (Note: If you're wondering why I did not include a $+C$ in this paragraph, it's because I never used an integral symbol. If you use an integral symbol, you are asking for all antiderivatives of a function. In this paragraph, I was simply stating one of the infinite number of antiderivatives these functions could have, in this case, where $C = 0$.)

You'll get into a weird situation with the power rule for integrals if you have something like

$\int \frac{1}{x}dx$

To solve, let's rewrite the integral.

$\int x^{-1}dx$

If we use the power rule, then we would get $\frac{x^0}{0}+C$, which is $\frac{1}{0} +C$, and we simply cannot work with that, because dividing by zero is a big no-no.

This is where the natural logarithm comes in.

$\int x^{-1}dx= ln|x| +C$

This should make sense because, when taking derivatives, you learned that the derivative of $ln(x)$ is $\frac{1}{x}$, which is $x^{-1}$.

One thing that you should note is the absolute value sign. Technically you should include that when solving integrals like these. You can think of it as a mechanism to prevent taking the natural log of a negative number. Remember: the inputs of $ln(x)$ can never be negative, and similarly, the outputs of $e^x$ cannot be negative.

Now, a common mistake that students have is this (don't actually write this down; it's incorrect):

$\int \frac{1}{x^2}dx=ln(x^2)+C$

The problem with this is that the natural log rule should not be applied. Rather, the power rule for integrals should be applied like this:

$\int \frac{1}{x^2}dx= \int x^{-2}dx = \frac{x^{-1}}{-1} + C = - \frac{1}{x} + C$

To avoid this mistake, remember that you should get the natural log as an answer only if $x$ has a power of $1$ in the denominator (or equivalently, a power of $-1$ in the numerator).