For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Introduction to Logarithms
 A logarithm is a way to reverse the process of exponentiation. It allows us to mathematically ask the question, "Given some base and some value, what exponent would we have to use on the base to create that value?"
 The logarithm base a of x (log_{a} x), where a > 0 and a ≠ 1, is defined to be the number y such that a^{y} = x.
log_{a} x = y ⇔ a^{y} = x.  The idea of a logarithm can be really confusing the first few times you work with it, so make sure to watch the video to clarify how logarithms are used and what they mean.
 The two most common logarithmic bases to come up are the numbers e and 10. As such, they have special notation because we have to write them so often.
 The base of e is expressed as ln. It is called the natural logarithm. [Remember, e is called the natural base.]
lnx ⇔ log_{e} x  The base of 10 is expressed with just log: if no base is given, it is assumed to be base 10. It is called the common logarithm.
logx ⇔ log_{10} x
 The base of e is expressed as ln. It is called the natural logarithm. [Remember, e is called the natural base.]
 Just like exponentiation, we can find the value (or a very good approximation) of a logarithm by using a calculator. Any scientific or graphing calculator will have ln and log buttons to take logarithms base e and 10, respectively. However, many calculators will not have a way to take logarithms of arbitrary bases. There is a way around this called change of base, and we'll explore it in the next lesson, Properties of Logarithms.
 When we graph logarithms, we see they grow very slowly. (This is because they are the inverse of exponentiation.) The graph of a logarithm also approaches the yaxis asymptotically. It gets very close, but it doesn't touch it.
 Logarithms are the inverse of exponentiation, and viceversa. The exponential function of base a is the inverse of the logarithmic function of base a:
f(x) = log_{a} x f^{−1}(x) = a^{x}  Because a logarithm is the inverse of exponentiation, we cannot take the logarithm of some numbers. Exponentiation (a^{x}) only has a range of (0,∞), so the corresponding logarithm can only reverse those same output values. In other words,
Domain of log_{a} x: (0, ∞) .  We can also see the above domain must be true because it would make no sense otherwise. Consider that log_{2} 0 = b means that 2^{b} = 0. But no such number b exists that could do this! The same goes for negative numbers, so the domain of any logarithm must be (0, ∞).
Note: This means we have now introduced a new type of thing to watch out for when we're looking for domains. Before, we only had to worry about dividing by 0 and having negatives under a square root. (Depending on what you've done previously, you might also have needed to be careful about certain things with trigonometric functions.) Now we also have to watch out for taking a logarithm of 0 or a negative number. One more thing to pay attention to when you're finding the domain of a function.
Introduction to Logarithms

 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to create the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  Thus, we can follow this swapping procedure to see an alternate, exponential way to write the equation given in the question:
[Notice that you can visualize it as if the base of the logarithm "slides" under the number that is opposite the logarithm.]log_{5} 125 = 3 ⇔ 5^{3} = 125
 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  With this definition in mind, we see that evaluating the logarithm is the same as figuring out what number the question mark would be below:
log_{4} 16 = ? ⇔ 4^{?} = 16  Therefore, all we need to do is figure out what exponent on 4 will create 16. Pretty quickly we see that 4^{2} = 16, so we have that log_{4} 16 = 2.
 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  With this definition in mind, we see that evaluating the logarithm is the same as figuring out what number the question mark would be below:
log_{2} 16 = ? ⇔ 2^{?} = 16  Therefore, all we need to do is figure out what exponent on 2 will create 16. We start working through exponents of 2:
Thus, we see that 2^{4}=16, so we have that log_{2} 16 = 4.2^{0} = 1 2^{1} = 2 2^{2} = 4 2^{3} = 8 2^{4} = 16
 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  With this definition in mind, we see that evaluating the logarithm is the same as figuring out what number the question mark would be below:
log_{25} 1 5= ? ⇔ 25^{?} = 1 5  Therefore, we need to figure out what exponent on 25 will create [1/5].
Clearly, a positive exponent will not allow us to create the fraction we're looking for. However, we can try negative exponents to make smaller values closer to 0.25^{0} = 1 25^{1} = 25 25^{2} = 625  Trying negative exponents out, we get:
Since [1/5] is somewhere between 1 and [1/25], we see that we need an exponent somewhere between 0 and 1.25^{0} = 1 25^{−1} = 1 2525^{−2} = 1 625  At this point, we hopefully remember that a fractional exponent has the same effect as a radical (see the lesson on the properties of exponents if this is unfamiliar to you):
Therefore, if we combine having a negative exponent with a fractional exponent, we can achieve the value we're looking for:a^{[1/n]} = n√a⇒ 25^{[1/2]} =
√25= 5
therefore log_{25} [1/5] = −[1/2].25^{−[1/2]} = 1 5,
 First, it's important to realize that ln is just a shorthand way of writing log_{e}. Thus, the question we're dealing with could be rephrased as
lne^{47} ⇒ log_{e} e^{47}  The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  With this definition in mind, we see that evaluating the logarithm is the same as figuring out what number the question mark would be below:
log_{e} e^{47} = ? ⇔ e^{?} = e^{47}  Once written out in this format, it's quite easy to see what exponent on e would create e^{47}: clearly, 47 is the only reasonable choice! Thus, we see that our logarithm must give that as the result.
lne^{47} = 47

 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  Thus, we can follow this swapping procedure to see an alternate, logarithmic way to write the equation given in the question:
[Notice that you can visualize it as if the base of the exponential expression "slides" out and becomes the base of the logarithm on the other side.]7^{5} = 16807 ⇔ log_{7} 16807 = 5

 The logarithm for the base a and some number x is defined as follows:
In other words, the logarithm of a number gives the appropriate exponent (for the log's base) to make the number (using that base) that the logarithm is acting upon.log_{a} x = y ⇔ a^{y} = x  Thus, we can follow this swapping procedure to see an alternate, logarithmic way to write the equation given in the question:
[Notice that you can visualize it as if the base of the exponential expression "slides" out and becomes the base of the logarithm on the other side.]e^{−4.2} = 0.014 995 … ⇔ log_{e} 0.014 995 … = −4.2  Finally, when writing log_{e}, we typically use the shorthand version written as ln instead. [log_{e} is still also correct, just less common.]
 When working out the graph of this function, remember that we can think of a logarithm as figuring out the question mark for the below:
With this in mind, we realize the below:log_{4} x = ? ⇔ 4^{?} = x log_{4} 1 = 0 log_{4} 4 = 1 log_{4} 16 = 2 log_{4} 64 = 3  When graphing any function, it helps to make a table of values so we can plot points. With logarithms, it makes it a lot easier for us if we choose our xvalues so the the logarithm will produce "clean" results, like we saw above. Working on our table, we find:
x f(x) 1 0 4 1 16 2 64 3 256 4  Now that we know what happens as we go to the right, what about as we go the left? At this moment, we need to realize that logarithms only have a domain of (0, ∞). [Check out the video lesson for a detailed explanation of why this is the case.] Because of this, the function simply does not exist for x=0 or less.
 However, the function does exist for 0 < x < 1. So what happens for those small fractions? Notice that [1/4] = 4^{−1}. Thus, log_{4} [1/4] = −1. This process continues, and we can add it to our table of values:
Thus, we see that the function has a vertical asymptote at x=0, shooting down to −∞ as x becomes very, very small.x f(x) 1/4 −1 1/16 −2 1/64 −3 1/256 −4  Finally, we're ready to put this all together and draw the graph. In choosing the axes, it is useful to choose a very large set of xvalues for a comparatively small set of yvalues because the function takes so long to grow. Plot points, then connect with curves.

 First, let's deal with the domain. (Remember, the domain of a function is the set of all input values it can accept.) For any logarithm, regardless of the base, it can accept the set of values (0, ∞). In this case, though, it's not simply logx, so the domain is not simply (0, ∞).
 Notice that the logarithm is acting on (−x+5), so what we need to do is figure out when the quantity (−x+5) outputs something in the set (0, ∞). That is, when is the below inequality true?
(−x+5) > 0  This inequality is true when
so we have a domain that is x: (−∞, 5).x < 5,  Now that we know the domain, we need to figure out the range. (Remember, the range of a function is the set of all values it can output.) If we look at the graph of a standard logarithmic function (such as the one we worked out in the previous question), we see that the function can output values of (−∞, ∞). [Why is this the case? A logarithm is basically a way of figuring out what the value of the question mark in an expression like 10^{?} = number. Since there exists a number for any value you choose to plug in for the question mark, it's possible that the question mark can be any value. Thus, it's possible for the logarithm to output any value.]
 With the specific function we're working on in this problem, this range will still hold true. While it's not precisely a standard logarithmic function because the logarithm is acting on (−x+5), we haven't affected the values that the logarithm will output in any way. Thus, the range is (−∞, ∞).

 First, let's deal with the domain. (Remember, the domain of a function is the set of all input values it can accept.) For any logarithm, regardless of the base, it can accept the set of values (0, ∞). In this case, though, it's not simply lnx, so the domain is not simply (0, ∞).
 Notice that the logarithm is acting on (x−19), so what we need to do is figure out when the quantity (x−19) outputs something in the set (0, ∞). That is, when is the below inequality true?
(x−19) > 0  This inequality is true when
so we have a domain that is x: (19, ∞).x > 19,  Now that we know the domain, we need to figure out the range. (Remember, the range of a function is the set of all values it can output.) If we look at the graph of a standard logarithmic function (such as the one we worked out in the previous question), we see that a normal logarithmic function can output values of (−∞, ∞). However, for the function we're working on, that won't be the case. But it's a useful starting place to keep in mind.
 Notice that the logarithm is being squared: [ln(x−19) ]^{2}. Thus, while ln(x−19) can output (−∞, ∞), once it is squared, it can no longer produce negatives. Thus [ln(x−19) ]^{2} can only output [0, ∞). Next, notice that the function then adds 7 to that:
Therefore, the output of [ln(x−19) ]^{2} is "raised up" by 7, resulting in a range of [7, ∞) for g(x).g(x) = ⎡
⎣ln(x−19) ⎤
⎦2
+ 7
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Introduction to Logarithms
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Introduction 0:04
 Definition of a Logarithm, Base 2 0:51
 Log 2 Defined
 Examples
 Definition of a Logarithm, General 3:23
 Examples of Logarithms 5:15
 Problems with Unusual Bases
 Shorthand Notation: ln and log 9:44
 base e as ln
 base 10 as log
 Calculating Logarithms 11:01
 using a calculator
 issues with other bases
 Graphs of Logarithms 13:21
 Three Examples
 Slow Growth
 Logarithms as Inverse of Exponentiation 16:02
 Using Base 2
 General Case
 Looking More Closely at Logarithm Graphs
 The Domain of Logarithms 20:41
 Thinking about Logs like Inverses
 The Alternate
 Example 1 25:59
 Example 2 30:03
 Example 3 32:49
 Example 4 37:34
Precalculus with Limits Online Course
Transcription: Introduction to Logarithms
Hiwelcome back to Educator.com.0000
Today, we are going to have an introduction to logarithms.0002
At this point, if we want to find the value of a number raised to an exponent, it is easy.0005
We use our exponentiation rules and evaluate; if it is something simple, like 2^{3},0008
then we know that is 2 times 2 times 2, 2 times itself 3 times, which we figure out is 8.0012
And we also figured out how to do numbers that weren't just simple integer exponents.0018
But we have all of these nice rules from our previous few lessons.0022
But what if the question was inverted, and what if we knew the base and the end result, but we don't know the exponent that we need to get there?0025
If we knew that we had 2 as the base, and we wanted to have an end result of 8,0033
but we had no idea what exponent we had to use to get to 8, how could we figure out that exponent?0038
This is the question that leads us to explore the idea of the logarithm, which we will be looking at over the next few lessons.0045
We define the logarithm as a way to talk about this unknown exponent.0052
The logarithm base 2 of x, denoted log_{2}(x), with this little 2 as a subscript right here,0055
is defined to be the number y such that 2^{y} = x.0065
So, when we see log_{2}(x) = y, then we know that that would be 2^{y} = x.0071
In other words, we are looking for the number that...when the logarithm goes onto some number,0079
we are trying to figure out what value, if it raised to this space here, would become the number that we took the logarithm of.0085
So, for the example of log_{2}(8) = 3, the reason why that is the case is because0094
we are looking for what number here we have to raise to to get 8.0100
So, the answer there is 3; if we raise 2^{3}, we get 8, so we know that log_{2}(8) = 3.0105
It can be a little bit confusing to remember how this works at firstwhat the notation means.0115
So, when you see log_{2}(x) = y, you can think of this as...if the 2 were under the y,0120
if we had 2^{y}, it would make the x that we are taking the logarithm of.0129
So, we take the logarithm of some number in regards to some base.0135
And that tells us what number we would raise the base to, to get the original number we are taking the logarithm of.0139
It will make more sense as we see more examples.0146
Here are some other examples: log_{2}(32) = 5, because if we raise 2 to the 5^{th}, we get 32.0149
2 times 2 times 2 times 2 times 2 is equal to 32; 4, 8, 16, 32.0158
log_{2}(1) = 0 because, if we raised 2 to the 0, from our rules about exponents, we know that that is just the same thing as 1.0168
log_{2}(1/4) = 2, because we know that, if we raise 2 to the 2, then that is the same thing as 1/2 raised to the 2,0177
where you flip to the reciprocal; and 1/2 squared would become 1/4, which is what we initially started with.0186
So, a logarithm is a way of taking a logarithm of a number, so that you figure out0194
what you would have to raise some base to, to get the thing you took the logarithm of.0199
We can expand this idea to something beyond just base 2, to a general idea.0204
The logarithm base a of x, log_{a}(x) (that little a, right down here, is a subscript), where a > 0,0209
and a is not equal to 1 (our base has to be greater than 0, and our base has to not be equal to 1),0218
is defined to be the number y such that a^{y} = x.0224
If we take log_{a}(x), then we know that that gives us some y, and that a^{y} = x.0228
So, once again, it is the same idea, where, if we take this base, and we put it under the y, we would get a^{y}.0236
And then, we would have the thing that we were originally taking the logarithm of, which is what we have there.0245
That is what is occurring right here.0249
The idea of the logarithm is that you take the log, and it tells you something that you can raise a number to, to be able to get this other value.0251
It is a little bit complex the first time you get it; but as you do it more and more, it will start to make more sense.0260
And we are going to see a whole bunch of examples to really get this cleared up.0264
Now, notice: we have these restrictions on what our base can be.0267
We know that the base has to be greater than 0, and the base is not equal to 1.0271
The base a of a logarithm has the same restrictions as the base of an exponential function.0274
This is because exponentiation and logarithms are inverse processes; they do the opposite thing.0280
And we will see more about how they are inverses in the future.0288
But they do reverse things, so they have to have the same restrictions, because they are basically working with the same idea of a base.0291
They are being seen through different lenses, and it will make more sense as we work on it more and more;0299
but we have to have the same restrictions on itotherwise the idea of a logarithm will just not make sense, or not be very interesting.0303
So, we have these restrictions: that we have to have our base greater than 0, and we have to have our base not equal to 1.0309
Let's look at some examples to help clear this idea up.0316
log_{7}(49) = 2, because, if we move this base over,0319
then 7^{2} is just 7 times 7, which gets us 49, which is exactly what we started with.0324
So, this is the case, because 7^{2} = 49; or maybe let's write it in the way that we had it originally here: 49 =...0332
moving the base over, moving our base underneath the right side...we have 7^{2}, like this.0343
The same thing over here: we know that log_{10}(10000) = 4,0350
because, if we move our base under, we know that 10^{4} is equal to 10000.0354
The question is: if we want to know what number we have to raise 10 to, to get 10000that is what log_{10}(10000) is effectively asking.0363
It is saying, "What number do we have to raise 10 toraise our base toto get 10000 as the end result?"0379
10 to the what equals 10000? The answer to that is 4.0387
So, we take 10^{4}; we get 10000; and sure enough, we see that that is 10 times 10 (100) times 10 (1000) times 10 (10000).0391
The same thing is going on over here: if we have log_{5}(1/125), then we move that under, and we get 5^{3} = 1/125.0400
Let's check that out: 5^{3} is the same thing as 1/5 to the positive 3; we flip to the reciprocal.0413
And then, 5 times 5 times 5 is 25...125; so we get 1/125; sure enough, it checks out.0423
Finally, if we take log_{4}(2), then we see that 4^{1/2} is equal to 2.0433
What is 1/2? 1/2 is an exponent that means square root; 4^{1/2} is the same thing as √4, which is 2; so once again, that checks out.0441
It is the question of what exponent I am looking for to be able to get this base to become the number that I am taking the logarithm of.0451
We can even do this with more unusual bases on our logarithms.0459
For example, if we have log_{1/2}(1/16), then we can see that that will become 4, because (1/2)^{4} is equal to 1/16,0462
because 2, 4, 8, 16...1/2, 1/4, 1/8, 1/16; so we see that that is the same thing.0475
If we take log_{π}(π), then we know that it has to come out as 1,0486
because π^{1}of course, that is no surprise that it is going to equal what it already started with.0492
So, if we are taking log_{π}(π), then the thing that π has to be raised to, to get π, is just 1.0499
So, we have 1 as the thing that comes out of that.0505
If we take log_{√2}(4√2), then we get that that has to be 5, because (√2)^{5}sure enough, that is equal to 4√2.0509
We can check that out: √2 times √2 times √2 times √2 times √2.0524
Well, √2 times √2 becomes 2; √2 times √2 becomes 2; and we have this one √2 left.0531
So, 2 times 2 is 4, and we are left with 4√2; it checks out.0537
The final one: log_{e}(1) becomes 0, because if we move this over, e^{0}, just like anything raised to the 0, becomes 1.0542
So, it checks out; so we have some idea of how it works.0553
A logarithm is, "For this base, what number do I have to raise it to, to get the number that I am originally taking the logarithm of?"0555
When I take the log_{10}(10000), it is a question of what number I have to raise 10 to, to get 10000.0562
I have to raise 10 to 4 to get 10000.0570
When I take log_{7}(49), it is a question of what number I need to raise 7 to, to get 49; the answer to that is 2.0574
That is the idea of a logarithm.0582
The two most common logarithmic bases to come up are the numbers e (remember, e is the natural number;0585
we talked about it previously, when we talked about exponential functionsa very important idea) and 10.0591
As such, they have special notation, because we have to write them so often.0597
The base of e is expressed as ln; so when we want to talk about base of e, the shorthand for that is ln.0601
It is called the natural logarithm; remember, e is called the natural base.0607
So, when we are taking a log_{e}, we call it a natural logarithm, and we use ln, because originally,0612
the French were the ones who came up with this; so it was logarithme naturel (excuse my FrenchI am not very good at speaking French).0617
So, the natural log of x is equivalent to log_{e}(x); ln(x) is just a shorthand way of saying log with a base of the number e.0623
Base 10 is expressed with just log on its own; notice, it has no subscriptthere is no little number down there.0634
If no base is given, it is assumed to be base 10; since base 10 comes up a lot, it is just an easy way to write it; this is normally what it means.0641
It is called the common logarithm, because it is a commonlyused logarithm.0648
So, if you see log(x), notice that it has no little subscriptno little number down there.0652
Then, we know that that is going to mean log_{10}(x).0657
Well, we can find the value of expressions like log_{2}(8); we know that that came out to be 3,0662
because the number that we raised 2 to, to get 8, is 3.0667
How do we figure out the value of more complicated expressions?0670
Like if we wanted to figure out the natural log of 12.19and as we just saw, that would be the same thing as asking, "What is log_{e} or 12.19?"0673
Well, e is a very complicated number; it goes on foreverit is irrational.0681
12.19 is not a very friendlylooking decimal number; so how are these two things going to interact?0685
We can guess that it is probably not going to come out very cleanly, in a nice way.0690
Sure enough, it doesn't: it comes out to be approximately 2.500616; and precisely, it would keep going forever, as well.0693
So, for calculating logarithms, just like exponentiation, we can find the expressions, or a very good approximation, by using a calculator.0701
Any scientific or graphing calculator will have natural log and log_{10} buttons to take logarithms base e and 10, respectively.0709
However, many calculators will not have a way to take logarithms of arbitrary bases.0718
So, if we had log_{3}, most calculators won't have an easy way for us to just get what log_{3}(some number) is.0723
But there is a way around this, and it is called change of base.0731
So, if you do need to take the log base 3 of some number, check out the next lesson, Properties of Logarithms,0734
where we will explore how you can change from one base to another.0740
So, the way that you calculate complicated logarithms like this is: you generally just use a calculator.0743
The calculator has a way, a method, to be able to figure out what that comes out to be.0748
Now, just like with exponentiation that we talked about before, we should note that there are ways to calculate these values by hand.0753
We could do this by hand and figure it out; and you will learn about this in more advanced math classes.0759
But we won't learn about it in this course right here.0764
Doing this takes a lot of arithmetic, though; and so we designed calculators to speed up the process.0766
It is something that we could do; it is not like we are completely reliant on calculators for figuring this idea out.0771
Logarithms weren't something that we only got once we had calculators created.0776
We have been able to have this idea for a very long timesince the 1600s, in fact.0780
But being able to calculate what these numbers come out to bethat takes a long time; it is a slow process.0785
So, we have calculators to be able to figure this out for us very quickly and very easily.0790
So, it speeds things up, but it is not that we are dependent on calculators.0794
It is just that they are a useful tool that we can apply in this situation.0797
Graphs of logarithms: so now, since we can evaluate logarithms however we want,0802
because we have these nice calculators as tools, we can plot graphs of them.0806
So, let's look at some graphs: f(x) = log_{2}(x) (this is in red); g(x) = log_{5}(x)0809
(that one is in blue), and finally h(x) = log_{10}(x) (that one is in green).0817
Notice how short the yaxis is; it only goes from 3 up until positive 5.0823
But we go all the way out to positive 100 on the xaxis.0830
We can see that hereright here; it is hard to seethat is a 1 value on the xaxis.0835
And that is going to end up corresponding to 0, because log of anythinglog of any base of 1will come out to be 0,0842
because the number that you raise anything to, a^{0}, = 1.0855
So, if we would take log base anything of 1, it is going to always come out to be 0,0859
because that is the number that we raise anything to, to get 1 in the first place.0867
So, that is why we see a common height of 0 there.0871
And notice how slowly they grow: at 16, log_{2} is only going to be at a meager 4.0873
But for log_{10}, when we look at log_{10}, it takes getting all the way up to 100 to even get to 2.0881
If we go out here to the 2, it takes all the way to 100 to be able to get that from log_{10}, because 10^{2} = 100.0889
We are seeing a similar thing for log_{5}: it has to get all the way up to 25 before it hits this height of 2, as well, because it is 5^{2}.0899
And we aren't even going to see it hit height 3, because it is not going to hit a height of 3 until it manages to get to 1250909
as an input value, because 5^{3} becomes 125.0915
So, notice how slowly these graphs grow.0919
These graphs grow really, really slowly, because for logs, it takes a really big number to be able to get even slight increases in our verticals.0923
And the farther out they go, the even slower they are going to grow.0930
Now, notice that they approach the yaxis asymptotically.0933
So, as they get smaller and smaller, they get really, really close to this yaxis right here.0937
They never touch or pass it, although that might be hard to see in this picture, since it looks like it is right on top of it.0943
But they get very close; they won't actually touch it, but they get very close to the yaxis.0948
We will talk about this behavior of how it gets really close to the yaxis,0954
and why it can't actually touch the yaxis or go past it, soon, when we talk about the domain of a logarithm.0957
The logarithm is the inverse process of exponentiation.0963
For example, let's consider base 2: if we have log_{2}(x) = y, then we can see its flip of 2^{y} = x.0966
We just change the x and the y location.0973
So, if we take log_{2}(8), that becomes 3, because remember: 2^{3} = 8.0975
So, when we take log_{2}(8), we get 3.0982
But then, if we take that 3, and we plug it into the other onewe take the 3, and we plug it in up here0985
we look at 2^{3}look: we are right back where we started.0990
We have the same thing on both sides.0993
We take this log, and we do something to it, and then we do the reverse process with the same base as the exponentiation.0997
We get back to the original input that we put in.1003
The same thing: we did it the other way, where we did exponentiation first.1006
If we take 2^{2}, then that is going to flip to (1/2)^{2}; so we would get 1/4.1009
And then, if we take log_{2}(1/4), we are going to get 2.1013
So, exponentiation and logarithms are doing inverses: one goes one way, and one goes the other way.1018
Together, they cancel out; we will be discussing this idea a lot more in the coming lessons.1023
It is a very important thing; we will also be proving it in general.1027
We can see this as one in a general form for any logarithm.1030
The exponential function of base a is the inverse of the logarithmic function of base a.1034
It is critical, though, that they do have the same bases; our exponential function is base a, and our logarithmic function must be base a.1039
If they are not the same base, it won't work.1046
Let's see why this is the case: if we have f(x) = log_{a}(x), f^{1}(x) = a^{x}.1048
Then, we can take f^{1}(f(x)) and see what happens.1055
Now, remember: we are talking about stuff from our lesson on inverse functions.1061
If you need more background on inverse functions, make sure you go and check out that lesson.1065
It will help you understand what is going on here.1068
So, f^{1}(f(x)) =...well, we could do this as...since this is a^{x}, then it is going to be...1071
well, we will apply f^{1} next; first, f(x) = log_{a}(x).1077
We have log_{a}(x); then we apply the f^{1}, and we have a^{loga(x)}.1082
Now, what does that end up coming out to be?1092
Well, remember: log_{a}(x) = y is the same thing as saying a^{y} = x.1094
So, that is what log_{a}(x) is: it is this y, some y such that if we were to put it as an exponent on a, we would get x.1104
So, log_{a}(x) = y: we can just say, "Whatever the number log_{a}(x) is, let's call it y."1115
So, we can swap that out, and we can say, "a^{y}," just because we are saying we will call log_{a}(x) y.1120
That is what we have over here; but remember, we defined this idea of what log_{a}(x) is based on a^{y} = x.1126
Well, we now have a^{y} = x; so if a^{y} = x, then that equals x,1133
which means that f^{1}(f(x)) = x.1139
Whatever we put in as our input comes out as our output if we do these two functions, one on top of the other.1144
We have inverse functions, because one function cancels out the effects of the other function.1149
We will talk about this more in future lessons.1154
We can also see this in the graphs of exponential and logarithmic functions.1157
So, if we take two graphs of, say, 2^{x} (that one is in red) and log_{2}(x), we see them like this.1161
And then finally, we also have y = x in yellow here, coming through the middle.1168
Now, remember from our lesson about inverse functions: when we learned about inverse functions,1174
we know that if two functions are inverses, they mirror over the line y = x.1180
They are swapping x and y coordinates; this shows us that they have to be inverses.1188
For example, if we look at what log_{2} at 2 is, it comes out to be a height of 1; here is a height of 1.1193
And then, if we look at what our 2^{x} at 1 is, at 1 it is a height of 2.1201
So, for this one, we have (I'll colorcode it back to what it had been) (1,2).1212
But for the blue one, we have (2,1).1223
They flip x and y locations, and that is going to be true wherever we go on this, because we see that they do this thing with y = x,1226
where they mirror across it; their x and y locations swap, showing us that they are inverses.1236
Notice all the graphs that we have seen of logarithms: they never pass, or even touch, the yaxis.1243
They never pass the yaxis; they never even manage to touch the yaxis.1248
This is because the domain of a logarithm is 0 to infinity.1252
And notice: there is a parenthesis on that 0; so it says it is not inclusive1257
so, not including 0, everywhere up from 0 (but not including 0), all the way up to positive infinity.1260
We can see this for a couple of reasons.1266
First, since exponentiation and logarithms are inverses, that means that the range of an exponential function is the domain of a logarithm.1268
The range of f(x) = a^{x} is going to be 0 to infinity.1275
a^{x}...if we put in any base a that is greater than 0 and not 1, it is going to go anywhere from 0 up until infinity.1280
If we look at 2^{x}, by varying what we plug in for x, we are going to be able to get anything between 0 and positive infinity.1288
Now, let's talk briefly about this idea: if we had a pool of numbers that we called a, the set of things that we are allowed to use,1297
and then we had another pool of numbers that was b, the set of things that it is possible to get to through some function f...1306
we have some function f that takes numbers from a, and it goes to b; then we call the numbers over here domain.1315
We talked about this when we first talked about the ideas of functions.1322
So, here is the domain of f; and over here is the range of f.1326
The domain of f is everything that f is able to take in; the range of f is everything that f is able to put out.1331
So, for the example a^{x} or the example 2^{x}, as a specific example,1338
the domain is anything; it can take in any number at allnegative infinity to positive infinityany real number whatsoever.1343
But it is only going to be able to give out numbers from 0 to infinity.1352
So, in this case, we see that it is going to have its range as 0 to infinity.1355
Now, notice: if we do the reverse of this, if we want to see the reverse of this, a function that does the opposite of what f does,1359
f^{1}, then it is going to have to go, not from a, but from b.1368
So, its domain, the domain of f^{1}, is going to be going in the other direction.1373
Since it is taking what f did and reversing it, it has to be able to take the things that f does as outputs.1380
Whatever f makes as outputswhatever f puts outis what f^{1} will take in.1385
So, the domain of f^{1} is the range of f, which means that the range of f^{1} is also the domain of our original function, f.1391
f goes from a to b; f^{1} goes from b to a.1402
Now, we saw that, for any exponential function, its range is 0 to infinity.1407
So, that means that the domain of f^{1}, the domain of a logarithmic function,1412
since it is the inverse of exponentiation, must also be from 0 to infinity.1419
So, that is why we have this domain here; the domain of any log has to be from 0 to infinity,1425
because the range of any exponential function is from 0 to infinity.1430
So, they are going to be done as opposites; the range of an exponential function is the domain of a logarithmic function.1434
So, that is a fancy way to be able to understand why this has to be the case,1441
because we can say what we learned about inverse functions applies here, because we have an inverse function.1446
But alternatively, we can just see that it would not make senseit just is nonsense if we look at it otherwise.1450
Consider if we tried to take log_{2}(0); then we know that that has to be equal to some number b for it to be a possible thing.1456
Then that means that 2^{b} has to somehow be equal to 0.1463
But that doesn't make any sense: no such number b exists.1470
No possible number could exist that would be able to take 2 and turn it into 0.1474
2^{b} can't ever become 0; if we plug in any number, we can make very small numbers; but we can't actually get all the way to 0.1480
We can't touch 0; the same is going to go for negative numbers.1486
If we wanted to say 2^{b} = 4, there is no number that does that.1489
We can't raise 2 to some number and make it negativeit started out positive, so we can't possibly make it negative.1495
So, this is impossible; this is impossible; this is impossible; so it means that log_{2}(0) is an impossible idea.1501
We can't take the logarithm of 0 or anything that is going to be negative,1509
because it just won't be possible for it to work over here, where we are trying to figure out what would be the exponential version of it.1513
So, since it just doesn't make sense to take the logarithm of a number that is 0,1519
or to take the logarithm of a number that is negative, it must be that the domain is always positive.1523
We have to go from 0, but not including 0, all the way up to positive infinity.1528
We can take in any of those things, but we can't take in 0; we can't take in negative numbers.1533
That explains why our domain has to be this.1537
We can think about it that way, or we can think about it as this flipped idea of the fact that exponentiation and logarithms are inverses.1540
So, we can have this more complex idea of the domain and range of what those things have to be.1546
But we can also just go to the fact that it does not make senseit would be nonsense; that is a reasonable idea, too.1552
All right, we are ready for some examples.1559
Let's evaluate these numbers without a calculator.1561
If we are looking at log_{6}(216), then that is going to be equal to some number,1564
such that, when we raise 6 to that number, we get 216.1568
216 = 6^{?}: we want to figure out what this is, so let's see.1572
What are some numbers that we could get out of this?1577
6^{1}that is just 6; 6^{2}well, that would be 36; 6^{3} is 180 + 36 = 216.1579
That is what we are looking for; so it must be the case that it is 3: log_{6}(216) = 3; that is our answer.1588
If we have log(1/10000), the first thing we want to do is remember: if we have just log, then that is a way of saying it is log base 10.1600
So, log_{10} (1/10000)once again, we are asking what that is going to be.1609
Well, that is going to be the number such that 10 to whatever that number is is going to be equal to 1/10000.1613
So, let's look at possible numbers for 10; if we go positive, we have 10^{1} = 10.1624
Well, that is not going to work, because we are going to need a fraction.1629
So, we notice that 10^{1} is 1/10; and then, if we think about that, 10^{1} would be 1/10;1632
10^{2} would be 1/100; 10^{3} would be 1/1000; 10^{4} would be 1/10000.1638
So, 10^{4} = 1/10000; we can also see this, because we can count the number of 0's: 1, 2, 3, 4.1645
So, that is 10^{4}, and since it is 1/10^{4}, then that must be 10^{4}.1655
We have that 4 is what we have to raise 10 to, to get 1/10000.1661
The natural log of e^{17}: well, remember: natural log is just another way of saying log base e.1668
So, log_{e}(e^{17}): what number do we have to raise e to?1673
e^{?} =...well, the thing we are working with is e^{17}, so e^{17} would be e^{?}.1683
Well, that is pretty clear: the thing that the question mark has to be is the 17.1691
Otherwise they will never match up; so it must be e^{17} that we want here, so 17 is our answer,1695
because if we raise e to the 17, it is no surprise that we get e^{17}.1702
Finally, log_{4}(32): once again, we are saying, "What is the number that we have to raise 4 to, to get 32?"1707
So, we move that over; we can think of this as 4^{?} = 32.1714
So, 32 = 4^{?}: well, let's start looking at some possible numbers for 4.1720
We could have 4^{1}; that would just be 4not big enough.1725
4^{2} would be 16; we are starting to get close.1730
4^{3} would be 64it looks like we overshot.1733
Well, we might notice that 16 times 2 equals 32; so if we could somehow get 2 to show up, we would be good.1738
Notice: how is 4 connected to 2what is the connection between these numbers?1746
Well, the square root of 4 is equal to 2; but we also had another way of saying that: 4^{1/2} is the same thing as saying square root.1753
4^{1/2} = 2; so we see that 4^{2} times 4^{1/2} equals 32.1762
4 squared times 4 to the 1/2 (4 squared is 16; 4 to the 1/2 is 2)16 times 2gets us 32.1770
So now, we just need to combine those: 4^{2} times 4^{1/2} is just another way of saying 4^{4/2} times 4^{1/2}.1778
We can add themthey are on a common base; so, 4^{5/2}...1790
So, the answer for thisthe number that we have to raise 4 to, to get 32, is 5/2.1795
All right, what if we were doing the other directionif we wanted to write an exponential equation in logarithmic form?1803
We have these exponential equations: 3^{4} = 81and now we want to do it in the logarithm form.1809
Remember, we have that log_{a}(x) = y is the same thing as saying a^{y} = x.1814
Remember, our base herewe can think of it as popping up under what is on the other side of the equation.1824
So, this over here is the exponential form; this here is the logarithmic form.1830
Logarithmic form is this log stuff, and exponential is this a to the something stuff.1838
So, what we have is exponential forms here; we want to flip it.1843
3^{4} = 81: that is going to be log...what is our base? Our base here is a 3, so log_{3}...1847
what is the number that we are raising to? That is the blue, so we are not going to use that.1858
Finally, the number that we have is x...so, log_{3}(81) = the number that we have to raise to, 4, because 3^{4} = 81.1862
If we ask what number we have to raise 3 to, to get 81, that is going to be 4; 3^{4} = 81.1878
We can do this with any of this stuff: 10^{2.4} = 251.18.1887
Then, that is going to be...our base is 10, so we can write that as just log, because if we don't have a base, it just says log base 10.1894
log of what number? Our number that we are going to get to is 251.18,1902
and it actually keeps going, so we will leave those dots there to show that it keeps going.1908
And that is going to end up equaling 2.4, because the number that we have to raise 10 to,1915
to get 251.18, is 2.4, as was shown to us in our original exponential form.1921
So, another one: our base here is e, so we can write that as natural log of this number.1929
We were told that it comes out to be 4.1937
Finally, our base is π: so log_{π}(this number) = √(1/2), because we know that,1942
if we raised π to the √(1/2), we would get 2.2466, and continuing on.1955
So, that is how we were able to figure out that log_{π}(2.2466...) must be the square root of 1/2.1962
Graph f(x) = log_{3}(x): to do this, we want to start with a nice table to figure out the values.1971
x; f(x); notice that we probably don't want to just toss in numbers right away.1978
If we plug in 10, well, I don't know what number we have to raise 3 to, to get 10.1986
That is going to be something complicated; we have to use a calculator.1992
But we do know what it is going to be if we plug in numbers like, say, 3.1994
If we plug in 3, what number do you have to raise 3 towhat is log_{3}(3)?1998
What number do we have to raise 3 to, to get 3?2004
That is easy: we just have to raise it to the 1nothing at all.2006
We don't have to raise it to anything, other than what is already there; so just something to the 1 is what it starts as.2009
What about 9?well, what number do we have to raise 3 to, to get 9?2015
We have to square it, so we have to raise it to the 2.2019
We can keep going in this pattern: what number do we have to raise 3 to, to get 27?2021
We have to raise it to the 3.2025
What number do we have to raise 3 to, to get 81? We have to raise it to the 4.2027
And we can keep going if we want.2031
What if we went in the other direction? Well, for 2, we don't know what number we would have to raise 3 to.2032
But for 1, yes, we do know what number we would have to raise 3 to.2037
3 to the what equals 1? Just like everything else, 3 to the 0 equals 1.2039
We could go to 1/3: what number do we have to raise 3 to, to get 1/3? 1.2047
What number do we have to raise 3 to, to get 1/9? 2.2053
And it would get lower and lower and lower, the closer we got to 0.2057
Once again, we will never actually be able to get to 0, because there is no number that we could raise 3 to, to get 0.2060
But we can get really, really close to 0.2065
So, at this point, we are ready to actually try plotting it.2067
Notice: our xvalues go pretty widely; so let's look at xvalues going from 10 up to +100.2069
And let's look at our yvalues: our yvalues, our f(x) values, don't really manage to change very much.2078
So, we will look at yvalues only going from 3...oh, let's make it 5...up to positive 5.2083
OK, let's start drawing that in; we start here; here is our xaxis and yaxis.2092
Make a scale; the scale for the x will be in chunks of 10, because we have to cover a lot of ground: 10, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.2103
We can keep going if we want, but that is good enough for us.2120
So, here is a 10; here is 100; I will mark 50 in the middle; 1, 2, 3, 4, 5....50, and 10.2123
So, we can see the scale on it here.2131
For our verticals, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.2133
Here is 1 and 5, positive 1 and positive 5.2144
Great; now we are ready to plot down some points.2150
We see a 3; we are at 1; so we are very close here; we are a little under 1/3 of the way up to the 10.2152
So, let's say it is about here; at 9, just a little bit before the 10, we are at 2.2159
At 27 (10, 20, 30a little bit before that, but a little under 1/3 of the way towards the other side), we are at 3.2167
At 81 (100, 90, 80just a hair in front of 80), we manage to be at 4.2180
There we go; now we want to go the other way, as well.2190
At 1/3...at 1, we are at 0, so we are really, really close to that yaxis already; at 1/3,2194
now we are getting pretty close; we are at 1 at 1/9; we are practically on top of it,2202
but we will never actually be on top of it; we will just get really, really close.2206
And we can see that this pattern is going to continue: 1/27, 3; 1/81, 4.2210
So, as it gets really close to the 0, it is going to just shoot down really quickly.2216
So, let's draw this side in; it approaches this asymptotically.2221
It gets really, really close, but it will never actually touch it; the part where it looks like it is touching it is just my human error at fault.2228
But it is not going to ever quite touch it on a perfect graph.2235
It might look like it, because of the thickness of the lines, but it will never actually do it.2239
And as it grows more and more, it slows down, because it has to go even farther out to be able to get any growth.2242
It slows down the farther out it gets; and we graph log_{3}(x).2250
Cool; finally, what are the domains of these functions? f(x) = log_{7}(x + 2).2254
Remember: the idea we had was log_{a}(stuff); then this stuff here must always be positive.2261
So, it must be positive; otherwise, it just doesn't work.2271
If we try to take the log of 0, it doesn't work; if we try to take the log of a negative number, it doesn't work.2276
You always have to take the log of positive numbers, whatever the base is.2282
For any base, this is going to be the case; so it doesn't matter if it is base 7 or base fifty billion.2286
It is going to be the case that we have to have whatever is inside of the logarithm,2290
whatever the logarithm is operating uponit has to be greater than 0; it has to be a positive number.2294
So, we know that the thing that log is operating on here is x + 2.2300
So, we know that x + 2 must be positive; it must be greater than 0.2304
We move the x over; we have that 2 has to be greater than x, so x has to be less than 2,2308
and it can go all the way down to negative infinity, because the only restriction we have is that 2 is greater than x,2314
which we could write out as...anywhere from negative infinity up until positive 2, but not including positive 2,2318
which we show with a parenthesis to show that it is not inclusive.2325
Over here, g(t) = 5t(log_{π}(3t + 7)).2329
Once again, the base doesn't really matter; it has to be positive, no matter what the base is.2333
For any arbitrary base a, it has to be positive on what the logarithm is operating on.2339
We look at the 5t part: we might get worried"oh, is the 5t going to interact with it?"2345
5t times log_{π}...5t is really in its own world; it is doing its own thing.2349
5 times t...we can do that for any number; we can multiply 5 times any number, so its domain is anything at all.2353
It is not going to actually get in our way; once again, the only thing we are worried about is2359
when the logarithm is going to try to take the log of a negative or 0 number.2362
So, to avoid that, we have to have that 3t + 7 must be greater than 0;2368
otherwise, we will be taking the log of something that we cannot take logs of, that would break our function.2373
So, 3t + 7 is greater than 0; subtract 7; 3t > 7; divide by 3; t must be greater than 7/3.2378
So, t starts at 7/3, but is not actually able to include 7/3.2388
So it starts just above 7/3 and can go anywhere larger; so it can go all the way out to positive infinity.2392
So, we have 7/3 shown with a parenthesis, because we can't actually include 7/3; we can just get arbitrarily close to it.2397
It is going all the way out to positive infinity.2404
And there are our two domains; all right, cool.2407
We will talk a bunch more about logarithms in the next one, where we will explore the properties of logarithms.2409
And then, we will see even more about how the two connect.2413
We have a lot of really interesting ideas; it is new stuff, but once you start practicing it,2414
as you do it a bunch of times, logarithms will really start to make sense.2418
You will get this idea of "what am I trying to raise this number to, to get the thing I am taking the logarithm of?"2420
What does this base have to be raised to, to get the number that I am taking the log of?2426
All right, we will see you at Educator.com latergoodbye!2430
2 answers
Last reply by: Duy Nguyen
Thu Aug 6, 2015 10:24 PM
Post by Duy Nguyen on August 6, 2015
Hi, would you mind explaining why the domain of a log function does not include negative numbers? Because log base (2) of (8) is 3 and x, in this case, is a negative number.
Thank you very much.