Lesson Video: Graphs of Logarithmic Functions Mathematics

In this video, we will learn how to sketch logarithmic functions with different bases and their transformations, and we will study their different characteristics.

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Video Transcript

In this video, we’ll learn how to sketch logarithmic functions with different bases and their transformations, and we’ll study their different characteristics. So, let’s begin by reminding ourselves what we actually mean when we call a function logarithmic.

Now, a logarithmic function is the inverse of an exponential function. It’s of the form 𝑓 of π‘₯ is equal to log base 𝑛 of π‘₯, where 𝑛 is greater than zero and not equal to one. And we say if the point π‘₯, 𝑦 satisfies the exponential function, then the point 𝑦, π‘₯ satisfies the logarithmic function. And the fact that the exponential and logarithmic functions are inverses of one another is actually really helpful when it comes to sketching their graphs. And we will investigate that in a little more detail later on in this video. For now, let’s begin by investigating the shape of the curve given by a logarithmic function.

Find the missing table values for β„Ž of π‘₯ equals log base two of π‘₯.

And then we have a table with three missing values. So, what do we mean when we talk about a logarithm? Logarithmic functions are inverses to exponential functions. Imagine we have the given expression. We read this as log base 𝑏 of π‘Ž equals 𝑐. 𝑏 is the base, 𝑐 is set to be the exponent in this expression, and π‘Ž is called the argument. And in fact, this is exactly the same way as describing the relationship between π‘Ž, 𝑏, and 𝑐 as 𝑏 to the power of 𝑐 equals π‘Ž. So with this in mind, let’s take the function log base two of π‘₯, and we’ll take our first value of π‘₯, negative two.

Substituting π‘₯ equals negative two in the function for β„Ž of π‘₯ gives us β„Ž of negative two is equal to log base two of negative two. But we need to work out the value of β„Ž of negative two. So let’s call that 𝑐 sub one. If we equivalently describe this relationship as two to the power of 𝑐 sub one equals negative two, we see we need to find the value of 𝑐 sub one that satisfies this equation. But there is no power of two that will give us a value of negative two. This could in fact only be achieved if the base itself was negative. So, 𝑐 sub one is in fact not defined. β„Ž of negative two then, we say, is undefined.

Let’s now move on to π‘₯ equals one. β„Ž of one in our table will be the value of log base two of one. Defining β„Ž of one to be equal to 𝑐 sub two this time, we can equivalently write this relation as two to the power of 𝑐 sub two equals one. To solve this equation, we ask ourselves, what power of two gives an answer of one? Well, in fact, the only way for this to be true is if 𝑐 sub two equals zero. Any real number not equal to the power of zero raised to the power of zero will always be one. So β„Ž of one and the second value in our table is zero.

We’re now going to repeat this for π‘₯ equals two. β„Ž of two is log base two of two. Defining β„Ž of two as 𝑐 sub three, we see that we can represent this relation as two to the power of 𝑐 sub three equals two. So once again, we ask ourselves, what power of two gives two? Well, the only power of two that will give us two itself will be one. So 𝑐 sub three, and in fact β„Ž of two, is equal to one. And so we’ve completed the values in our table; they are undefined, zero, and one, respectively.

We might now look to sketch the graph of this function. And to do so, we’ll need to work out a few more values. And we could, of course, use the same method as before. Alternatively, we could simply type these into a calculator. Let’s suppose π‘₯ is equal to four. Log base two of four is equal to two. Similarly, if π‘₯ is equal to eight, we get log base two of eight, which is equal to three.

Now, β„Ž of negative two is undefined. In fact, the function itself is undefined for any values of π‘₯ less than or equal to zero. So, we have an asymptote here, an asymptote of the line π‘₯ equals zero or the 𝑦-axis. So, the graph of our function β„Ž of π‘₯ equals log base two of π‘₯ looks a little bit as shown. The domain, we said, is the values of π‘₯ we can substitute in, and that’s purely positive real numbers: π‘₯ is greater than zero or π‘₯ is contained in the open interval from zero to ∞. We also observe that the graph itself passes through the π‘₯-axis at π‘₯ equals one. And we can generalize these two properties for graphs of logarithmic functions.

Consider graphs of the form 𝑦 equals log base 𝑛 of π‘₯, where 𝑛 is greater than zero and not equal to one. They have just one π‘₯-intercept. They pass through the π‘₯-axis at one. In fact, they will also pass through the point 𝑛, one. In our last example, the graph of 𝑦 equals log base two of π‘₯ passes through the point two, one. These graphs all have a vertical asymptote given by the 𝑦-axis or π‘₯ equals zero. Finally, these functions have a domain where π‘₯ is in the open interval zero to ∞ and a range where the output, the 𝑦-values, are in the open interval from negative ∞ to ∞. So it looks a little something like this.

Now, in fact, just as with exponential graphs, the size of 𝑛 can tell us a little bit more about the function. If 𝑛 is greater than one, then the function itself is increasing. And if 𝑛 is greater than zero or less than one, the function is decreasing. Now in fact, we might even look to compare the graph of this function with its inverse, 𝑦 is equal to 𝑛 to the power of π‘₯. The graph of 𝑦 equals 𝑛 to the power of π‘₯ looks a little something like this. As we might expect when we draw the graph of a function and its inverse, they will be reflections of each other in the line 𝑦 equals π‘₯. Now this can be a really useful fact if we’re struggling to remember what either graph looks like. So now let’s have a look at a question that involves identifying the graph of a logarithmic function.

Which graph represents the function 𝑓 of π‘₯ equals log base five of π‘₯?

And we have five graphs to choose from. So, let’s begin by inspecting the function we’ve been given. It’s a logarithmic function, and it has the general form of the logarithmic function 𝑓 of π‘₯ equals log base 𝑛 of π‘₯ where, of course, our value of 𝑛, which is here five, cannot be equal to one and is greater than zero. Now, one of the features we know about this function is that it passes through the π‘₯-axis at one, but it also passes through the point 𝑛, one. So we’re looking for a graph which passes through one, zero and five, one. We also know that the 𝑦-axis or the line π‘₯ equals zero is an asymptote to this graph. In other words, the graph of our function approaches the 𝑦-axis but never actually reaches it. And we know that when 𝑛 is greater than one, the graph itself is purely increasing. It’s increasing over the entire domain of the function.

Well, in fact, all of our graphs, if we look carefully, are increasing and they all have the 𝑦-axis as an asymptote. So we need to identify which of our graphs pass through the point one, zero and five, one. In fact, if we plot each of these points on every single graph, we see that only one graph passes through both points. The correct graph, then, is (A). Graph (A) represents the function 𝑓 of π‘₯ equals log base five of π‘₯.

Let’s now identify the graph of a second logarithmic function.

Which curve is 𝑦 equals log base three of π‘₯?

And then we have a coordinate plane with four individual graphs drawn on it. Well, here we have a logarithmic function, a function of the form log base 𝑛 of π‘₯, where 𝑛 is positive and not equal to one. One of the properties of this function is that its graph passes through the point one, zero and the point 𝑛, one. So our graph, since 𝑛 is equal to three, will pass through one, zero and three, one. We also know that if our value of 𝑛 is greater than one, then the graph is increasing over its entire domain.

And this is really useful because we can actually disregard options (c) and (d) straightaway. We notice that those graphs, whilst they do pass through the point one, zero, are decreasing over their entire domain. And so they could be a transformation via reflection of a logarithmic graph or a logarithmic graph with a fractional base. But they certainly aren’t the curve of 𝑦 equals log base three of π‘₯. Finally, we know that all of these curves should have an asymptote given by the line π‘₯ equals zero or the 𝑦-axis. Well, all of our graphs do have this asymptote. We observe that the graphs seem to approach the 𝑦-axis but never quite reach it.

So, we need to find the curve then that passes through the point one, zero and three, one. We have already shown that all curves pass through the point one, zero, but the only curve that also passes through the point three, one is curve (a). So, curve (a) is the graph of the equation 𝑦 equals log base three of π‘₯.

So at this point, we’ve considered how to generate a group of ordered pairs using a table of function values, and we’ve identified a couple of logarithmic graphs. It might be clear, though, that we won’t always be working with logarithmic functions on their own. Sometimes we might be dealing with composite functions or a function of a function. In these scenarios, it can be helpful to recall what we know about function transformations to help us identify the relevant graph. Let’s demonstrate one such transformation in our next example.

Which function represents the following graph? Is it (A) 𝑓 of π‘₯ equals log base four of two π‘₯? (B) 𝑓 of π‘₯ equals log base four of π‘₯. Is it (C) 𝑓 of π‘₯ equals log base two of two π‘₯? Is it (D) 𝑓 of π‘₯ equals log base two of π‘₯? Or is it (e) 𝑓 of π‘₯ equals log base eight of π‘₯?

We’ll begin by inspecting the graph of the function itself. It certainly resembles the graph of a logarithmic function. We have a vertical asymptote of the 𝑦-axis. The graph certainly seems to approach this line but never quite reaches it, and the function itself is increasing. The domain is values of π‘₯ greater than zero and the range appears to be all real numbers. And so it’s likely that our function is of the form 𝑓 of π‘₯ equals log base 𝑛 of π‘₯. And, in fact, all of the equations that we’ve been given are of this general form, although some of them have two π‘₯ as their argument.

So one of the things we know about the function 𝑓 of π‘₯ equals log base 𝑛 of π‘₯ for positive values of 𝑛 not equal to one is that they pass through the point one, zero and the point 𝑛, one. By plotting the point one, zero on the graph of our function, we see that it doesn’t actually pass through this point. In fact, it passes through the point one-half, zero, so it looks like it could be a transformation of our original function. We might then recall that say we have a function 𝑦 equals 𝑓 of π‘₯, the function 𝑦 equals 𝑓 of 𝑛π‘₯, where 𝑛 is not to be confused with the base in log base 𝑛 of π‘₯, gives a horizontal compression but by a scale factor of one over 𝑛.

So, if we had the function 𝑓 of π‘₯ equals log base 𝑛 of π‘₯, the function 𝑓 of π‘₯ equals log base 𝑛 of two π‘₯ would be a compression by a scale factor of one-half in that horizontal direction. So, this means that our graph could be (A) or (C). It could be log base four of two π‘₯ or log base two of two π‘₯. Let’s choose a point on the curve to deduce whether it’s 𝑓 of π‘₯ equals log base four of two π‘₯ or 𝑓 of π‘₯ equals log base two of two π‘₯. We see our curve passes through the point eight, two and the point two, one. For due diligence, we’ll check both of these coordinates in our individual functions.

When π‘₯ is equal to two, the first function, (A), gives us 𝑓 of two equals log base four of two times two. That’s log base four of four, which we see is equal to one. So, this point satisfies the function 𝑓 of π‘₯ equals log base four of two π‘₯. Then, 𝑓 of eight is log base four of two times eight. That’s log base four of 16. And we know four squared is equal to 16, so this must be equal to two. Once again, our second point eight, two satisfies this function, so we can deduce that the function is 𝑓 of π‘₯ equals log base four of two π‘₯.

Let’s check, though, by substituting two and eight into our second function. When π‘₯ is equal to two, the function is log base two of four, which is two. And when π‘₯ is equal to eight, the function is log base two of 16, which is equal to four. Neither point two, two nor eight, four lie on the graph we’ve been given, and so we can confirm that it cannot be option (C); it’s (A).

We’ve now demonstrated how to identify and sketch graphs of logarithmic functions as well as looked at their various transformations. So, let’s finish up by recapping some of the key points from the lesson. In this lesson, we saw that the logarithmic function is the inverse of an exponential function. It’s of the form 𝑓 of π‘₯ equals log base 𝑛 of π‘₯, where 𝑛 is positive and not equal to one. Their graphs pass through the point one, zero and 𝑛, one. They have an asymptote given by the 𝑦-axis or the line π‘₯ equals zero, and the value of 𝑛 can tell us about its shape. If 𝑛 is greater than one, the graph is increasing over its entire domain. And if 𝑛 is between zero and one, it is decreasing.

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