If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra (all content)>Unit 11

Lesson 30: Graphs of logarithmic functions (Algebra 2 level)

# Vertical asymptote of natural log

What happens to ln(x) as x approaches 0? Created by Sal Khan.

## Want to join the conversation?

• At , Sal says that "this is only defined if you're taking the natural log of a POSITIVE value" (my emphasis). I don't see that as being obvious. Why can't this be defined by taking the natural log of a negative value? Can you please direct me to the right videos in this site? Many thanks! This site is wonderful.
• The reason this is so is because of this logic. log 100 = 2 (given that we are using log base ten) so another way to write this is 10^2 = 100. Makes sense right? If not make sure to clear up that understanding before reading on. So let's say we have log (-100). What does that equal? There is NO SOLUTION! Think of solving (10^ x = -100), there is no REAL number (let's stick to real numbers since the imaginary ones get a little complex :) ) that x can be to solve this. Since 10 is already positive you can't take it to any power to make it negative. If you have a graphing calculator notice that the whole left side of the equation is blank if you plug in y = 10^x OR log y = x (same thing).
• This question is somehow off-topic but I couldn't find a proper place to ask: We already know that there is something called imaginary number `i^2 = -1`. Why don't we apply it here? Sal said `ln (x-3) defined for x > 3`, is it possible for us to say `ln (x -3) defined for x < 3 also if we use i`?
• You have correctly observed that we could use complex numbers for the logs of negative numbers (but not of 0).

The question is why do we skip over this material in beginning calculus? Because it sometimes can be pretty difficult math and you have enough to be getting on with just learning the basics of limits and differentiation.

The basic property of a log of a number that is not positve is this:
Let k > 0. The log's base is, of course, e.
log (−k) = log k + π 𝑖
log (-k 𝑖) = log k − ½ π 𝑖

That isn't too hard, but when you start working with full complex numbers such as Log (3 + 4𝑖) it can get very tricky (getting into something called branches and branch cuts). I won't go into that, it is not easy.

So, we generally just gloss over logs of negative numbers in beginning calculus. I'm not so sure that is a good idea, but that is how it is often done.

However, log 0 is undefined for any base.
• At , what property/rule is Sal using to take both sides of an equation and treat them as exponents to base 'e'? He does: e^(ln (x-3)) = e^0
How generalizable is this property? For example, if I have the equation: 2x =0, can I do e^2x = e^0 ?
• As long as you don't do something that is or might be undefined or otherwise not allowed by the rules, if you do something to both sides of the equation you maintain equality.

However, when working with roots, logarithms and sometimes exponential functions, you can get extraneous answers.

For example, in the problem you mentioned:
2x = 0, this has only one solution
e^(2x) = e⁰, this has infinitely many solutions, but only x=0 is a valid solution for the original problem, the rest are extraneous.
Just for reference, e^(2x) = e⁰ = 1 has the following solutions:
x = n*i*π where n is any integer. But only when n=0 and thus x=0 is this a valid solution for the original equation.
• No horizontal asymptote?
• No, the logarithms have no horizontal asymptote, they continue to grow to infinity on the positive side, and they are limited by the vertical asymptote on 0.
• I've typed into my calculator ln(-1) and it gave me pi i. You can try with google. It's interesting, does anybody know why is that?
• Sure, you have an advanced calculated that can handle complex numbers.
While it is usually taught in earlier math courses that the log of a negative number is undefined, that is not true. Here is the actual solution:

let k be any number greater than 0.
ln (−k) = ln (k) + π𝑖
Thus,
ln (−1) = ln (1) + π𝑖
• There are asymptote that cross over the curve many times. But a asymptote is defined to be line that when infinitely extended , the distance between curve and line approaches zero. How is it possible when it has crossed the curve earlier and distance has earlier became zero?
(1 vote)
• Look at the graph of f(x) = 10 sin (5x)/x, as an example. This should make it clear how this happens.
• Two bits of terminology I haven't seen before: y(x) and natural log. Can you direct me to previous videos. Thanks.
(1 vote)
• What did Sal do in ? I can't figure out what he's doing raising both sides of the equation to e and then simplifying the left side and e disappears! Can anyone explain in easy-to-understand English?
• Question on semantics here. Is there something different between "taking the natural log" and simply multiplying by the natural log?
• "Taking the natural log" refers to a mathematical operation, similar to saying "taking the derivative" or "taking [x] to the power of [y]."

I think you may be confused here by the notation. In math speak, "taking the natural log of 5" is equivalent to the operation ln(5)*. You're not multiplying "ln" by 5, that doesn't make sense. The ln symbol is an operational symbol just like a multiplication or division sign. If you said "five times the natural log of 5," it would look like this: 5ln(5). If you're still confused, I'd recommend you go to the logarithms section in the Algebra II playlist.

*What this actually signifies is the number you need to raise Euler's Number (notated by "e," roughly 2.718281) to to get 5. So you could also say that ln(5) means "e to what number equals 5."
• Is this function defined only for positive x-values because no exponentiation of any number gives zero or a negative result?
(1 vote)
• If you have moved onto calculus, then you need to know that the logarithm IS defined for negative values of x. You are taught in Algebra II or other basic math course that it is not defined in order to avoid having to deal with nonreal numbers.

But, so you know:
given that k > 0
ln (-k) = ln (k) + iπ

So, you will be needing to learn to work with logs involving complex numbers.

However, ln(0) is undefined. The natural log is actually defined by a limit and that limit fails to exist for x=0:
ln (x) = lim h→0 {xʰ - 1}/h

There is obviously a singularity at x=0, which is why ln(0) fails to exist.