Logarithmic Functions - Online Tutor, Practice Problems & Exam Prep

Logarithms Introduction

Hey, everyone. Whenever we solve polynomial equations like x3=216, we can simply do the reverse operation in order to isolate x. So, here, since x is being cubed, I could simply take the cube root of both sides to cancel that, isolating x, giving me my answer that x=216. But what if my variable x is instead in my exponent, like this 2x=8? How are we going to isolate x here?

Well, I know that, looking at this, I can just think, okay, how many times does 2 need to be multiplied by itself in order to give me 8? And I know that my answer is simply x=3. But what if I'm given something like 2x=216? I really don't want to have to multiply 2 enough times to get up to that 216. So, is there not just an operation I can do to cancel out that 2 and leave me with x?

Well, here I'm going to show you that there is an operation that does just that because the reverse or inverse of an exponential is actually taking the logarithm, or a log. Now, the first time you work with logs, they can be a little bit overwhelming. But here, I'm going to walk you through exactly what a log is and how we can use it to actually make our lives easier, especially when working with exponents. So now that we know that the reverse or inverse operation of an exponential is simply taking the log, we can go ahead and take the log of both sides of this equation in order to isolate x. So, I take log(2x), and that's equal to log(216).

But we actually do need to consider one more thing here because whenever we canceled the 3 on our x3, we took the cube root. We didn't take the square root or the 4th root or anything else. We took the cube root to cancel that 3. We need to consider something similar when working with logs because logs and exponentials need to have the same base as each other to cancel. So here, since I have an exponential of base 2, I want my log to have a base of 2 as well.

So really, I want to take a log base 2 of 2x and then, on the other side, a log base 2 of 216. Now that my log and my exponential have the same base, then it's going to fully cancel out, leaving me with just x=log2(216). Now it's fine to leave it in this form here. This is actually called our logarithmic form. And we're later going to learn how to fully evaluate these and get a number, but for now, we're just going to keep it in that log form.

Now that we're here, what exactly does this statement mean? We have log2(216). Well, a log is actually giving us the power that some base must be raised to in order to equal a particular number. But what does all of that mean? Well, looking at our function here or our equation here, log2(216), this is really saying, okay, what power does 2 need to be raised to in order to give me 216?

Now, this statement here I mentioned is in its logarithmic form, and it's actually an equivalent statement to our very first equation, 2x=216. This is just in its exponential form, and we basically translated it into its logarithmic form. Now we're going to have to do this for multiple statements. We're going to have to translate and convert expressions between these two forms. So diving a bit deeper in converting between these two forms, let's start by taking this equation in its exponential form, 3x=81, and putting it into log form.

Now whenever we convert between these two forms, no matter what we're going to or from, we're always going to start at the same place. We're going to start with our base. So this base 3 of our exponent is going to become the base of our log. So I start here with log3. Now once I have my base, I'm going to circle to the other side of my equal sign to that 81.

So I have a log3(81), and then once I have that 81, I'm going to circle back to the other side of my equal sign, and that's what log3(81) is going to equal, so that's equal to x. Now I have my equivalent statement in its log form. log3(81) equals x is equivalent to 3x=81. Now that we've seen going from exponential to log form, let's go in the reverse direction from log to exponential form. So starting with this statement here, x is equal to log4(64), we're going to start at that same place.

We're going to start with our base. So here, my log has a base of 4. That becomes the base of my exponential. so I have a log4 and I have that 4 and now I want to raise 4 to a power and I'm going to raise it to the power that is on the other side of my equal sign. So start with your base, go to the other side of your equal sign.

In this case, I get x here, so 4x, and then finally circling back to the other side of our equal sign, that is our final number here, 64. So 4x=64 is an equivalent statement in exponential form from this x equals log4(64). Now I know that that might seem like a lot right now, so let's just walk through some examples together. So let's start with this x is equal to log5(800). Since that is in its log form, I have a log right there, I wanna go ahead and put this in exponential form.

So remember, we're going to start at the same place every single time no matter what we're starting with. We always start with that base. So here I have log5, so I'm going to start with an exponential of base 5. Now once I have that base, I'm going to circle to the other side of my equal sign and get that x. So 5x.

And then I'm gonna circle back to the other side of my equal sign, and this is equal to 800. So start with your base, other side of your equal sign, circle back to where you started. So 5x=800 is my equivalent statement in its exponential form. Let's look at another example here. We have log2(16) equals 4.

Now where are we going to start here? Well, we wanna start with our base, of course, so I have this log2, so that becomes the base of my exponent. So I have 2, and then I go to the other side of my equal sign, 2x, and then circle back to where you started. 24=16. Now this is great because we can actually see that this is a true statement.

24=16, so I know that I wrote that correctly. Let's look at one final example here. Here we have 10x=45100. So this is in its exponential form. We want to go ahead

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 logb2⁡x. So let's start by just plugging in some values of \( x \) and getting some ordered pairs. So starting with \( x = 1 \), if I plug that into my function, I get logb2⁡1. And I know any time 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 \( x = 2 \) to my function, I get logb2⁡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 2nd 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 \( f(x) = 2^x \), 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 noticed that these are the same exact points with \( x \) and \( y \) flipped. Now 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 and -2), having flipped those points, there's my new point. And then for (1/2), 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: (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, \( y = x \), 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 \( y = x \). 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. Now 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 \( y = 0 \) for our exponential function, which then corresponds to a vertical asymptote now at \( x = 0 \) for our logarithmic function. Now let's consider one final thing here and consider when we have to graph logs of different bases. Now 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 1/2, our graph is instead going to be decreasing. Now that we know everything that we need to about graphing log functions, let's get some more practice.

Graphing Logarithmic Functions Example 1

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 logarithm 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. 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. So, starting 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 values. 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 negative one, \( \frac{1}{3} \), 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 it's going to get really steep on this side. And then on my left side here, it's 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 the x-axis right at \( y = 0 \). Now that we have our graph for \( 3^x \), let's move on to plotting \( g(x) \) where we have \( \log_3(x) \).

Now remember that whenever we're dealing with inverse functions, we can simply swap our \( x \) and \( y \) values. So these values that I have 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. So starting with this point, I'm going to go ahead and switch that up.

So now it is \( (1, 9, -2) \), just reversing my \( x \) and \( y \). And then for \( -1 \), \( \frac{1}{3} \), that becomes \( (\frac{1}{3}, -1) \). \( 0,1 \) flip is \( (1, 0) \). Then I have \( (1, 3) \), which I flip to get \( (3, 1) \). 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 \( g(x) \) to get our final function up here. So let's go ahead and start with this 10 and work our way down this way first. So 10, and then I have \( (3, 1) \), and then \( (2, 9) \) 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. 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 \( x = 0 \). So 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.

The Natural Log

Hey, everyone. We just learned about a frequently occurring logarithm, log base 10, and there's actually another log that occurs rather frequently, log base \(e\), called the natural log. Now don't worry. We're not going to have to learn how to do anything new here. Here, I'm going to walk you through how we can treat log base \(e\) just as we would any other log of any other base, just with a special notation.

So let's go ahead and get into it. Now log base 10, our common log, we would write as log, and similarly, log base \(e\) gets its own special notation. It's always written as ln. A way to remember that is because this is the natural log, if we take the first letter of each of those words, natural log, and we reverse them, we get ln, which we're always going to read out as natural log. Now the natural log also has its own special button on your calculator, the ln button, for whenever you need to use that to evaluate something on your calculator.

We just saw that whenever we have an exponential equation \(b^x = m\), we can rewrite this in its log form as log base \(b\) of \(m\) is equal to \(x\). Now the same thing is true with base \(e\). We treat it just like any other base. So if I have \(e^x = m\), I can rewrite that in its log form as log base \(e\) of \(m\) is equal to \(x\).

One thing we just need to be careful of here is that log base \(e\) should always be rewritten as simply ln. You're pretty much never going to see log base \(e\) written out like that. It's always going to be written as that ln. So with that in mind, let's walk through these examples together. Now our first example here is \(x\) is equal to the natural log, ln, of 17.

Let's go ahead and actually rewrite this as log base \(e\) to help this make a little bit more sense to us. So here we have \(x\) is equal to log base \(e\) because that's what the natural log is of 17. And we want to rewrite this in its exponential form because it's currently a log. So we're going to start at the same place we always do with that base. So here our base is \(e\).

Starting with that base and then circling to the other side of my equal sign, \(e^x\), and then circling back to my original side of my equal sign is equal to 17. And that's my final answer here, \(e^x = 17\), just like we would any other base. So let's look at one final example here. We have \(e^x = 4\), and we want to rewrite this in its log form.

Again, we're going to start at the same place with our base. So here, this base of \(e\) becomes the base of my log, so log base \(e\) of 4, and then circle back is equal to \(x\). Now we're not quite done here because remember we need to make sure we have the correct notation going on in our final answer. So we know that log base \(e\) is really just ln, the natural log. So this is really ln of 4 equals \(x\), and here's my final answer here.

So now that we know a little bit more about the natural log, let's get a bit more practice. Thanks for watching, and let me know if you have questions.

Change the following logarithmic expression to its equivalent exponential form.

$\log_4x=5$

• A. $4^{x}=5$
• B. $x^4=5$
• C. $4^5=x$
• D. $5^4=x$

Change the following logarithmic expression to its equivalent exponential form.

$x=\log9$

• A. $x=9$
• B. $9^{x}=10$
• C. $1^{x}=9$
• D. $10^{x}=9$

Change the following exponential expression to its equivalent logarithmic form.

$3^{x}=7$

• A. $\log_37=x$
• B. $\log_73=x$
• C. $\log_3x=7$
• D. $\log7=3^{x}$

Change the following exponential expression to its equivalent logarithmic form.

$e^9=x+3$

• A. $\log\left(x+3\right)=9$
• B. $\ln\left(x+3\right)=9$
• C. $\ln9=x+3$
• D. $\log_9x=e^3$