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.

$$e^x = 5 \\ ln(ex) = ln(5) \\ x = ln(5) \\$$

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?

$$ln(ab) = ln(a) + ln(b) \\ ln(a/b) = ln(a) - ln(b) \\ ln(a^b) = bln(a)$$

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:

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

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

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

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

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