Graphing Logarithmic Functions - Online Tutor, Practice Problems & Exam Prep

Graphing logarithmic functions is similar to graphing exponential functions since they are inverses. For example, the log base 2 of x can be derived by flipping the x and y values of its corresponding exponential function. Transformations such as vertical shifts and reflections can be applied to the parent function. The domain of a log function is from its vertical asymptote to infinity, while the range is all real numbers. Understanding these concepts is crucial for mastering logarithmic functions and their properties.

Graphs of Logarithmic Functions

Hey, everyone. Earlier when graphing exponential functions, we found and plotted a bunch of different ordered pairs, and we found that our graph had similarities to both polynomial and rational functions as we approached an asymptote. Now we're faced with graphing another type of function, a logarithmic function, and you may be dreading having to learn how to graph yet another different type of function. But I have great news for you here because graphing a log function is actually super similar to graphing an exponential function because a log is the inverse of an exponential. So here, I'm going to walk you through graphing a log function using exclusively things we already know about graphing exponential functions. So let's go ahead and get started.

Now looking at the log function we have here, we have log2(x). So let's start by just plugging in some values of x and getting some ordered pairs. Starting with x equals 1, if I plug that into my function, I get log2(1). And I know anytime I take the log of 1, no matter what the base is, I'm always going to get 0. So I have my first ordered pair at (1,0), which I can go ahead and plot on my graph. Then if I plug in x equals 2 to my function, I get log2(2), which since my base is the exact same as what I'm taking the log of, I know that my answer here is going to be 1, and I get my second ordered pair at (2,1).

Now let's pause here and take a closer look at these two points. I have (1,0) and (2,1). Now looking over to my exponential function here, I have f(x)=2x, which is the inverse function of this log function that we are trying to graph here. Now, looking at my exponential function, I have these two points, (0,1) and (1,2). And looking back to the points that we just found, I notice that these are the same exact points with x and y flipped. This is because these are inverse functions, so this is actually going to work for every single x and y value of our exponential function. If we simply flip them, we'll have all of the new ordered pairs needed to graph our log function. So let's go ahead and flip our x and y values and get these points plotted on our graph.

So looking at our first point here, (1/4, -2) having flipped those points, there's my new point. And then for (1/2, -1), I know f(x)=-1. Now flipping these final two points here, I'll end up with (4,2) and (8,3). And we can go ahead and plot all of these on our graph. So (4,2), (8,3), and then for our small fractions, (1/2, -1) and (1/4, -2). Now connecting all of these points, I have a complete graph of my log function.

And I notice here on this side that we're getting really close to that y-axis but not quite touching it, which tells us that we're, of course, dealing with an asymptote that I can go ahead and plot using a dashed line right along that y-axis to form my vertical asymptote.

Now that we have the complete picture of the graph of our log function, let's compare it to our original exponential function. Now looking at the graph of my exponential function here, if I were to take my entire graph and fold it along this diagonal line and then stamp my exponential function onto the other side, I would end up with a graph identical to the log function that we just graphed because this is actually exactly what we did. We simply reflected the graph of our exponential function over this diagonal line, y equals x, in order to get the graph of our log function. Now this will work for any log function and its corresponding inverse exponential function because they're inverses. They're always going to be reflections of each other along the line y equals x.

Now if you want a way to remember the shapes of these graphs, remember when you're dealing with an exponential function, you can think of a lowercase e. And if we just extend the tail of that e out, it looks really similar to the shape of the graph of our exponential function. And for a logarithmic function, if we look at our word logarithmic and we take that lowercase r and extend that r out, we get a shape really similar to the graph of our logarithmic function. So remember, for an exponential function graph, the shape looks really similar to extending a lowercase e, whereas for a logarithmic function, we extend that lowercase r.

Now that we have a good idea of what's going on with the shapes of these graphs and what they look like, let's also consider their domain and range. When graphing our log function, we were able to flip the x and y values of the inverse exponential function, and we can actually do the same exact thing to our domain and range. So the domain of our exponential function corresponds to the range of our inverse log function, so it's going to be all real numbers for any log function of any base. Then also the range of our exponential function corresponds to the domain of our log function, and it's going to depend on our asymptote and, in this case, go from 0 to infinity.

Now speaking of asymptotes, we also can take a look at our asymptote here, and we had a horizontal asymptote at y equals 0 for our exponential function, which then corresponds to a vertical asymptote now at x equals 0 for our logarithmic function. Now let's consider one final thing here and consider when we have to graph logs of different bases. Everything that we've looked at so far has been kind of the opposite of working with an exponential function, but this thing is going to be the exact same and the direction of our graph is going to depend on the value of our base b just as it did for exponential functions and it's going to depend on it in the exact same way. So for values of b that are greater than 1, like 2 or 3 or 10 or whatever, our graph is going to be increasing. And for values of b between 0 and 1, like say 1/2, our graph is instead going to be decreasing. Now that we know everything that we need to about graphing practice.

Graphing Logarithmic Functions

Hey, everyone. In this problem, we're asked to graph 2 different functions, \( f(x) = 3^x \) and \( g(x) = \log_3(x) \). Now we can see that these functions are inverses of each other because the log is the inverse function of an exponential, and we're going to go ahead and graph both of them together.

So let's go ahead and start with \( f(x) = 3^x \) and find and plot some points here. So starting with \( x = 0 \), we can plug that into our function and get \( 3^0 \), which is just 1. Now if I plug 1 into my function, that's \( 3^1 \), which is just 3. And then \( 3^2 \) is going to give me 9. Now let's go ahead and plot these points on our graph. I'm going to start with 1, and then I have 3. 9 is completely off of my graph, actually. We know that it's going to increase really rapidly and get really steep there.

Now for my negative values of \( x \), we're going to see the same numbers but in fractions as we had with positive. So \( 3^{-1} \) is simply \( \frac{1}{3} \) and then \( 3^{-2} \) is \( \frac{1}{9} \) and we can see that we're getting smaller and smaller as we go into those negative numbers. So let's go ahead and plot those points at \( -1, \frac{1}{3} \) and then \( -2, \frac{1}{9} \), really close to my x-axis. So we can go ahead and connect all of these points that we have on our graph here. I know that I'm going to get really steep on this side. And then on my left side here, I'm going to get really close to my x-axis but not quite cross it because we have an asymptote there. So I can go ahead and plot my asymptote as well. I know that it's at that x-axis right at \( y = 0 \).

Now that we have our graph for \( 3^x \), let's move on to plotting \( g(x) = \log_3(x) \). Now remember that whenever we're dealing with inverse functions, we can simply swap our \( x \) and \( y \) values. So these values that I had for \( x \) over here are simply going to become my \( y \) values when working with my new function that is the inverse. And then my \( y \) values over here are going to become the \( x \) values of my inverse function.

So let's go ahead and swap all of those points so that we can get them plotted on our graph. Starting with this point, I'm going to go ahead and switch that up. So now it is \( (9, 2) \). \( -1, \frac{1}{3} \) that becomes \( \frac{1}{3}, -1 \). \( 0,1 \) flip is \( 1,0 \). Then I have \( 3,1 \), which I flip to get \( 1,3 \). And then my last point flipping that \( 2 \) and that \( 9 \) I get \( 9, 2 \).

Now let's go ahead and plot all of these points for our function at \( g(x) \) to get our final function up here. So let's go ahead and start with this \( 1,0 \) and work our way down this way first. So \( 1,0 \) and then I have \( 1,3 \) and then \( 9, 2 \) goes off of my graph yet again. And then going the opposite direction into my fractions, \( \frac{1}{3}, -1 \) is going to be right about here. And then \( \frac{1}{9}, -2 \), we're going to be getting really close to our y-axis this time. So we can go ahead and connect all of these points to form our graph. And then we're getting really close to that y-axis, which tells us that we're dealing with a vertical asymptote up on this side right on our y-axis at \( x = 0 \).

Now that we have plotted both of our functions, we can go ahead and determine our domain and our range. Now when dealing with our domain of our first function, \( f(x) = 3^x \), I can see that this is going to be any value of \( x \). It can be all real numbers. Now, we know that our domain of \( 3^x \) is going to match up with the range of our function \( \log_3(x) \). So I know that my range here is also going to be all real numbers, which I can verify by taking a look at my function here.

Now, for my range of my function \( 3^x \), I need to go ahead and take a closer look and identify where my asymptote is so that I can get an accurate range here. Now when I consider my range, I know that we're not going to go past that asymptote here. So my range is simply going to go from my asymptote at that \( 0 \) point all the way to infinity. So my range here is simply from \( 0 \) to infinity. Now we know that our range is going to flip and become the domain of our other function of our inverse function. So that tells me that the domain of \( \log_3(x) \) is going to be from \( 0 \) to infinity, which we can again verify by taking a look at our graph. We see that we can't cross this asymptote, so we start at that \( 0 \) point and go to everything positive.

Now that we've graphed these inverse functions, let's get some more practice. Let me know if you have any questions.

Transformations of Logarithmic Graphs

Hey, everyone. We just learned how to graph basic log functions using the fact that they're the inverse of an exponential function. But what if we have to graph a more complicated log function like this g of x we have here? We might be worried that this is where it's going to start to get more challenging and switch up from exponential functions, but you don't have to worry about that at all because we're going to do the same thing we've done a million times before and not do any calculations, just some transformations. So let's go ahead and jump right in. Now our function g of x here looks really similar to log₂(x), but just has a couple of extra things going on here, this one and this four, both of which represent transformations.

Now, just as a quick recap of the most common transformations, remember that if you have a negative on the outside of your function, that represents a reflection over the x-axis, whereas the negative on the inside of your function represents a reflection over the y-axis. Then, of course, we have h, which represents a horizontal shift, and k, which represents a vertical shift. Now, just as we did for exponential functions, when working with log functions, we are always going to want to graph our parent function first because the base of your log function isn't always going to be the same. So let's go ahead and do that for this function g of x that we have here and start with step 0 and plot that parent function first.

So here, we want to go ahead and identify our parent function. I mentioned that in g of x, it looks really similar to log₂(x) because that actually is our parent function here. So our parent function f of x is equal to log₂(x). We want to go ahead and graph three points here at 1 over b where b is our base, so in this case 1/2, -1. Then our second point will be at 10, which will be the same no matter what the base of your log function is. And then our last point will be at b. base 1. So b here is 2. So that last point is at 2, 1. Now, these are going to be the easiest points for us to plot. And they're going to serve as sort of test points as we graph our new function. So plotting those on my graph, I have one, two, -1 then 10, and then 2, 1. Now I can go ahead and connect those points to get my graph here. And then I can also go ahead and plot my vertical asymptote at x equals 0 in order to get my entire parent function all ready to go and ready to be transformed.

Now that we have that parent function, we can go ahead and start plotting our actual function g of x here starting with step 1 and shifting that vertical asymptote to x equals h. Now here looking at my function I have that h is this one because remember x - h. So here h is just positive one. So I can go ahead and shift my vertical asymptote right over to that one using, of course, a dash line because it is an asymptote. K. Now that we have finished step 1, we can move on to step 2 and identify whether or not there's a reflection happening. Now remember, reflection happens if there is a negative on the inside or outside of our function. And here, looking at my function g, I don't have a negative that got added, so I don't have to worry about that reflection. And I can go ahead and move on to the second part of step 2 and shift my test points by h and k. Now we already identified h as being positive 1 and looking over at my function here this negative 4 tacked on the end here represents k. So I'm going to shift my test points of my function here by 1 point to the right and 4 points down. So let's go ahead and do that starting with this first point. I'm going to go 1 to the right and 1, 2, 3, 4 down, ending up right here. Then my next point, 1 to the right and 1, 2, 3, 4 down, and my final point, 1 over to the right, and 1, 2, 3, 4 down. So I have my new points here shifted by h and k, and then I can go ahead and move on to the very last step in plotting my function, which is going to be to sketch my curve approaching the asymptote.

So here I can go ahead and connect all of my new points and then approach that asymptote on the bottom there. Now that we have our complete picture of our graph, we can go ahead and identify our domain and our range as well. Now remember, our range is always going to be the same for our log functions no matter what that base is or what other shifts and transformations are happening. It's always going to be all real numbers. So there's my range, all real numbers. And then, of course, my domain actually depends on my asymptote and which direction I'm approaching my asymptote from. So looking at my graph here, I'm approaching my asymptote from that right side. And when that happens, that means that my domain is going to go from my asymptote at h to infinity. Now here h is 1, so my domain is going to be from 1 to infinity. Now if my graph was approaching my asymptote from the left side instead over here, my domain would then go from negative infinity until I reach that asymptote at h. Now that we know how to graph log functions using transformations, let's get some more practice.

Graph the given function.

$g\left(x\right)=\log_2\left(x-1\right)-4$

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