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### 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?

- At0:40, 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.(7 votes)
- 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).(10 votes)

- 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`

?(7 votes)- 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.(6 votes)

- At2:15, 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 ?(3 votes)- 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.(5 votes)

- No horizontal asymptote?(3 votes)
- 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.(3 votes)

- 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?(2 votes)
- 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(4 votes)

- 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.(3 votes)

- Two bits of terminology I haven't seen before: y(x) and natural log. Can you direct me to previous videos. Thanks.(1 vote)
- Understanding Function Notation:

https://www.khanacademy.org/math/trigonometry/functions_and_graphs/function_introduction/v/understanding-function-notation-exercise

Natural Logarithms has a wide range of videos. Go to this page to find your choice:

https://www.khanacademy.org/math/algebra/logarithms-tutorial(5 votes)

- What did Sal do in2:17? 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?(3 votes) - Question on semantics here. Is there something different between "taking the natural log" and simply multiplying by the natural log?(2 votes)
- "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."(2 votes)

- 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.(4 votes)

## Video transcript

- [Narrator] Right over
here, we've defined y as a function of x, where y is equal to the
natural log of x minus three. What I encourage you to do
right now is to pause this video and think about for what x values is this function actually defined? Or another way of thinking about it, what is the domain of this function, and then try to plot this
function on your own, on maybe some scratch paper that you might have in front of you, and then we'll talk about,
and also think about, does this have any vertical
asymptotes, and if so, where? So, I'm assuming you've given a go at it, so first, let's think about
where is this function defined? So, the important realization
is this is only defined if you're taking the natural
log of a positive value. So, this thing right over here must be, this thing right over
here, must be positive, or, another way we could think about it, it's only going to be
defined for any x, such that, x minus three is strictly positive. Not greater than or equal
to zero, greater than zero. We haven't defined what
the natural log of zero is, so x minus three has
to be greater than zero in order to give a positive, in order to take the natural
log of a positive something, or if we add three to both sides, we get x is greater than three. So, it defined, defined
four, x greater than three. You could do this as the domain, the set of all real numbers
that are greater than three. Now, that out of the way, let's actually try to plot
natural log of x minus three. So, let me put some of this
graph paper right over here, and the first thing I
wanna think about is, well, let's just try to plot
some interesting points here, and the most obvious one is
what makes this natural log, what makes this entire
function equal to zero? When are we going to intersect the x-axis? So, let's just think about
that for a little bit. So, when is the natural
log of x minus three going to be equal to zero? Well, one way to think about this is to view these both as exponents and raise e to both of these powers. So, you could say that e to the
natural log of x minus three is the same thing as e to the zero, and, of course, if you raise e to whatever exponent you need
to get you to x minus three, that's just going to get
you to x minus three, and if you raise e to the zero, well, anything to the zeroth power, except possibly zero, that
one's under contention, or maybe not defined, e to
the zero is equal to one. This is just another way of saying, "Hey, look, if I wanna know what exponent "do I need to raise e to to get to zero," we know e to the zeroth
power is equal to one, so x minus three is equal to one. So, if I'm taking the natural
log of one, it'll be zero. So, x minus three is equal to one, add three to both sides, you get x equals four. So, we know that the .4,
zero is on this graph, so let me graph that. One, two, three, and four. So, that right over there is the point. X is four, and y is zero. Four minus one, four minus three is one, natural log of one is zero. We also know that this is only defined for x being greater than three, so let's just put a little dotted line right over here at x equals three. And we know that our
function isn't even defined, for x equals three, and in
any value, to the left of it. Well, let's think about what happens as we approach x equals three
from the right-hand side, and to do that, I'll
make a little table here. So, let me make a little table here, and put some x values here, and then let's just think about what our corresponding y value is. So, we could put in, so we already know that we get four, zero. Let's try out 3.1, 3.01, and 3.001 and see what you get. And you can imagine, from each of these, you're gonna subtract
three, so then the input into the natural log function
is gonna be .1, .01, .001, and so you're gonna have more
and more negative exponents, or powers, that you have to raise e to to get to those values, but to just verify that, let's actually get our calculator out. Let's get our calculator out, and let me go to the main screen. And so, let's take the natural log of three, of 3.1 minus three. 3.1 minus three. We get negative 2.3, and
I'll just round to the tenth. So, this right over here is negative 2.3. If we raise, if we take the natural log, natural log of 3.01, 3.01 minus three, we get to negative 4.6, once again,
just rounding, negative 4.6, and if we take the natural log, the natural log of 3.001, zero, one, minus three,
we get to negative 6.9. Negative 6.9, and just for fun, let's do one that's way more dramatic. So, let's take the natural log of 3., let's do one, two,
three, four, five, zeros, followed by one, minus three, and we get a fairly more
negative value right over here. So, as you see, as we're getting
closer and closer to three, we're getting more, and more,
and more negative values. Let me just plot this right over here, so this is negative one,
this is negative two, this is negative three,
this is negative four. So, when x is equal to 3.1, which is right about there,
we're at negative 2.3, which is right around there. When x is 3.01, which is really
hard to see right over here, we get to negative 4.6,
so it's way down here. So, our graph is gonna
look something like, our graph is gonna look something like, and my best attempt is
to draw it freehand, is gonna look something, something like, something like that. So, do we have a vertical asymptote? Absolutely. As we approach three from
values larger than three, from the right-hand side, our
function is plummeting down. It's unbounded. It's going down. Our value of our function is quickly approaching negative infinity. So, we clearly have a vertical asymptote. We have a vertical asymptote at, asymptote, at x equals three.