Solving Exponential and Logarithmic Equations - Video Tutorials & Practice Problems

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1

concept

Solving Exponential Equations Using Like Bases

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In working with exponential functions, we would evaluate our function for some given value of X by simply plugging that value in for X and ending up with an answer. But what if our function is already equal to something? Well, then we're faced with finding the value for X that will then make our statement true. And now that we're faced with solving a new type of equation, you may be worried that we're going to have to learn an entirely new method of solving here. But you don't have to worry about that because I'm going to show you how we'll end up just solving just a basic linear equation that we've solved a million times before by simply doing one thing to our exponential equation and rewriting each side to have the same base from there. It's literally just solving a basic linear equation. So let's go ahead and get right into it. Now, here I have the equation 16 is equal to two to the power of X. So here I'm looking for the value of X that will make two to the power of that equals 16. So we want to rewrite these sides to have the same base in this two. I know that I can't rewrite as the power of anything. So I'm going to leave that as two to the power of X. And then I want to rewrite 16 to have that same base of two. Now, I know that 16 is simply equal to two to the power of four. And now both of these sides have that same base of two. Now, from here, we can simply go ahead and take our powers and set them equal to each other. So setting our powers equal to each other will end up with four is equal to X. And I'm actually done here. I don't even have to solve for anything, but typically you're going to have to solve for X here since X is already solved four, I already have my answer that X is equal to four. Now let's go ahead and look at some more examples just to get a bigger picture of what exactly is going on here. So looking at this first example, I have, I have 64 is equal to two to the power of X. Now, it might not always be immediately obvious how you can rewrite your powers. So let break this down a little bit more. Now, two to the power of X, I'm not gonna rewrite that two as anything. It's just gonna stay as that. But 64 then needs to get rewritten as a power with base two. Now, I don't know exactly what power two needs to be raised to, to get that so I can break this down a little bit more. I know that 64 is equal to eight squared. And I also know that eight is just two cubed. So if I take eight, break it down into two cubed and know that that eight has to get squared. This now has a base of two. Now, two cubed to the power of two is really just two to the power of six. So I have successfully rewritten that side 64 as a base of two. Now, I know that two to the power of six is equal to that two to the power of X that I originally had. And here my bases are now equal to each other. They're both two and I can simply take my powers six and X and set them equal to each other. So six is equal to X and I don't have anything left to do here. I already have my answer that X is equal to six. That will make my statement true. Let's move on to our next example. Here we have five to the power of X plus one is equal to the square root of five. Now both of these already have this five. So I know that that's what my base needs to be. And this five, the power of X plus one is going to stay five to the power of X plus one. But I know that the square root of five can be rewritten as an exponent. So I can rewrite this as five to the power of one half. Now, both of my bases are the same here. They're both five. So I can go ahead and take those powers and set them equal to each other. So I end up with the linear equation X plus one is equal to one half. Now I can easily solve for X here by simply subtracting one from each side. And I end up with X is equal to one half minus one, which will give me negative one half. So I have my final answer here and I just solved a basic linear equation by rewriting like base. So let's look at our final example here we have 27 is equal to nine to the power of X. Now, looking at this first glance, you may think, OK, since I have nine to the power of X, I need to rewrite 27 with a base of nine, but 27 cannot be rewritten with a base of nine. So I need to get a little bit more creative here. Sometimes you're going to have to rewrite both sides in order to get them to have the same base. So here I know that nine is equal to three squared. So I can rewrite this side as three squared. And that's to the power of X. But I know that if I'm raised that power to a power, this is really just three to the power of two X multiplying those exponents. Now 27 I know is equal to three to the power of three. So now I have the same base on both sides of three and I can take those powers three and two X and set them equal to each other. So I end up with three is equal to two X. And in order to solve for X, here, I can go ahead and divide both sides by two. And I'm left with my final answer that X is equal to three halves. Now I know that if I plug that back into my equation, that would make my statement true. Now that we know how to solve exponential equations by rewriting each side to have the same base. Let's get some more practice.

2

Problem

Problem

Solve the exponential equation.

$4^{x+7}=16$

A

2

B

5

C

9

D

– 5

3

Problem

Problem

Solve the exponential equation.

$100^{x}=10^{x+17}$

A

17

B

34

C

8.5

D

0

4

Problem

Problem

Solve the exponential equation.

$81^{x+1}=27^{x+5}$

A

– 13

B

11

C

14

D

3

5

concept

Solving Exponential Equations Using Logs

Video duration:

5m

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Hey, everyone, we just learned that whenever we have an exponential equation like 16 is equal to two to the power of X, we can simply rewrite each side to have the same base and then set our powers equal to each other in order to get our final answer. But what if we're given an exponential equation? Like 17 is equal to two to the power of X? It's not exactly easy to rewrite 17 is a power of two. And don't worry, you're not going to have to figure out what crazy decimal that two needs to be raised to in order to get 17 because here we can simply solve this equation using logs. So here I'm going to walk you through how we can just take the natural log of each side and use properties of logs that we already know in order to get our final answer. So let's go ahead and get started. Now let's just walk through an example together and just go through the steps that we're going to need to follow in order to get our answer here for each of these exponential equations. So for our first equation, we have 10 to the power of X plus 64 is equal to 100. So starting with step one, the very first thing we want to do is going to be to isolate our exponential expression. So here our exponential expression is this 10 to the power of X. So I want that by itself on one side of my equation. So in order to do that, I need to go ahead and move this 64 over which I can do by subtracting 64 from each side. Now, when I do that, I'm gonna end up with 10 to the power of X is equal to 36. Now, from here, step one is done and I can go ahead and move on to step two. Now, step two is when we're going to determine what log we should take, we're either going to take the natural log or the common log. And here is how we choose if it has a base of 10. If our exponential expression has a base of 10, which here I have 10 to the power of X, I'm going to take the common log just log of both sides. Now, if instead it did not have a base of 10, I would instead take the natural log. These are the only two logs you'll use whenever solving exponential equations. So here, like I said, we have a base of 10. So I can go ahead and just take the common log of both sides. So doing that, I get log of 10 to the power of X and that's equal to log of 36. Remember whatever you do to one side you have to do to the other. So we are of course, taking the log of both sides. Now step two is done and we can go ahead and move on to step three where we're going to use our log rules in order to get X out of our exponent. So here X is in my exponent, but I know that using my power rule, I can go ahead and move that over. So moving that X to the front here, I end up with X times log of 10 and that's equal to log of 36. Now, step three is done, we have X out of our exponent and now we can move on to step four and go ahead and solve for X. Now, looking at this equation, I have X times log of 10 is equal to log of 36. And you may notice here that I have this log of 10 and since log is just log based 10, I know that this is just going to end up being one. So this is really just X times one, which is just X. So here I end up with X and that's equal to log of 36. Now, here we could be done. X is equal to the log of 36. This is an acceptable answer if you're not asked to approximate because the log or natural log of some number is a constant. So we really don't need to do anything else from here unless we're explicitly asked to. So step four is done. But if you're asked to approximate, we can go ahead and plug this into our calculator. So just type in log of 36 and I could end up with a final answer of 1.56. And I'd be done. Now, let's move on to our next example, we have three is equal to two to the power of X plus one. Now, we want to restart our steps starting from step one and we want to go ahead and isolate our exponential expression. Now, our exponential expression is actually already by itself here. So I don't need to do anything else. Step one is done. I can move on to step two. Now, in step two, we're either going to take the log or natural log. And here I do not have a base of 10. So I'm gonna go ahead and take the natural log of both sides. So I here I take natural log of three and that's equal to the natural log of two to the power of X plus one. Now, step two is done and I can go ahead and move on to step three and use my log rules to take X out of my exponent Now again, here, I'm going to use the power rule. So I'm gonna take this exponent of by two and pull it to the front of that natural log. So on this left side, I still have the natural log of three. But now my right side has become X plus one times the natural log of two. Make sure you pull your entire exponent when using the power rule. So from here, I've taken X out of my exponent step three is done moving on to step four, we want to go ahead and solve for X. So this looks a little bit here, but we can simplify this significantly and get X all by itself. So here I have this natural log of three and that's equal to X plus one times the natural log of two. But remember that the log or natural log of something is just a constant and I can treat it just like I would any other number. So I can go ahead and move this natural log to the other side by simply dividing by it like I would any constant. So dividing both sides by the natural log of two, I end up with the natural log of three divided by the natural log of two and that's equal to whatever I have left over here, which happens to be X plus one. Now, I have one final step here in isolating X getting it by itself. And that is subtracting one from both sides. So I am left with that canceling. And I have my final answer here that the natural log of three divided by the natural log of two minus one is equal to X. Now, this looks a little bit scary. But remember these natural logs are just numbers. So this is really just a bunch of constants natural log of three natural log of two and one. So this is an acceptable answer. But of course, if you're asked to approximate using a calculator, you can go ahead and plug this into your calculator to get a decimal. Now, whenever you do that, you're going to get a final answer of X is equal to 0.58. And we are completely done here. We have our final answer. Now that we know how to solve exponential equations using logs as well as using like bases. We can solve any exponential equation that gets thrown at us. Thanks for watching and let me know if you have any questions.

6

Problem

Problem

Solve the exponential equation.

$2\cdot10^{3x}=5000$

A

$x=3.40$

B

$x=10.19$

C

$x=0.0001$

D

$x=1.13$

7

Problem

Problem

Solve the exponential equation.

$900=10^{x+17}$

A

$x=-14.05$

B

$x=2.95$

C

$x=0.17$

D

$x=1.72$

8

Problem

Problem

Solve the exponential equation.

$e^{2x+5}=8$

A

$x=-1.46$

B

$x=-1.11$

C

$x=-0.22$

D

$x=1.39$

9

Problem

Problem

Solve the exponential equation.

$7^{2x^2-8}=1$

A

$x=0.51$

B

$x=\pm2$

C

$x=\pm2.83$

D

$x=2.23$

10

concept

Solving Logarithmic Equations

Video duration:

5m

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Hey, everyone, we just solved a bunch of different exponential equations. But what about logarithmic equations? Well, now that we're faced with solving yet another new type of equation, you may be worried yet again, that we're going to have to learn something brand new in order to solve these. But again, you don't have to worry about that at all because here I'm going to show you that there are only ever going to be two types of log equations that you run into two logs of the same base that are set equal to each other or a single log set equal to a constant. And both of them boil down to solving a basic linear equation like we've done a million times before. So let's go ahead and just jump right in. Now with exponential equations, we saw that whenever we had exponents of the same base, we could simply take those powers and set them equal to each other in order to get our answer. And we can actually do the same exact thing whenever working with log equations. So here we have these log base too. And since these have the same exact base, I can simply take what we're taking the log of. So X plus one and five and go ahead and set those equal to each other. So I'm left with just a basic linear equation that I can then solve to get X is equal to four. So whenever we have two logs of the same base, we can just take what we're taking the log of and set them equal to each other and solve for X. So let's look at a slightly more complicated example here, we have the natural log of X plus four minus the natural log of two and that's equal to the natural log of eight. Now, here, it might not be immediately obvious that we're going to be able to just set some stuff equal to each other. So let's take a closer look here. I know that these are all natural logs. So they are all logs of base E that same base. And here I can actually use my log properties in order to condense this a little bit more. So I can actually go ahead and use my quotient rule here since I have this subtraction happening. So I can go ahead and condense this into the natural log of X plus four over to and that's equal to the natural log of eight. Now that these have the same base I have this natural log on both sides, I can go ahead and set X plus 4/2, equal to eight in order to get the linear equation that I can go ahead and solve for X with. So this becomes X plus 4/2 and that's equal to eight. Now, we can just solve this basic linear equation by isolating X. So my first step here is going to be to multiply both sides by two and that will cancel leaving me with X plus four and that's equal to eight times two, which is 16. Now, one final step in isolating X here is going to be to subtract four from both sides, leaving me with X is equal to 12. Now, that's my final answer. And I'm done. All I had to do was make sure that both sides had the same base and then go ahead and set things equal to each other giving me a basic linear equation. Now let's look at our other type of log equation if I cannot rewrite it with like bases and I actually just have a single log like say log base to a four X and that's equal to a constant. So here you may be worried that this is where it's going to be complicated, but it's not because we're simply going to put this in exponential form and then solve from there. So let's go ahead and work through this example together. So we have log base two A four X and that's equal to five. Now, our very first step here is going to be to isolate our log expression, which here is log based to a four X. So it's already isolated and I'm already done with step one, I can move on to step two, which is the meat of this problem putting it in exponential form. Now remember when putting things in exponential form, we're always going to start with that base. So here we're going to start with that too and then circle to the other side of our equal sign. So raise that two to the power of five and then come back to the other side of our equal sign and set that equal to four X. So we've completed step two. This is in its exponential form, we can go ahead and move on to step three and actually solve for X. So here I can go ahead and simplify this exponent this two to the power of five. I know that if I multiply two by itself five times, I'm going to end up with 32. So this is 32 and that's equal to four X. Now here to isolate X I can just divide both sides by four, canceling on that side, leaving me with my answer that X is equal to eight. Now, here I already have my answer. I've completed step three, but we actually have to perform one final check here because we actually want to go ahead to our very last step and check our solution by plugging X into M now M is just what we are taking the log of. So in this case, it's gonna be this four X here. So we want to plug our answer into that four X. And in this case, since we got an answer of eight, we're gonna plug in eight for X. So I'm gonna get four times eight, which is simply 32. And the thing we're looking for here is the sign of this number, whether it's positive or negative. So here, if M is greater than zero, if it's positive, we are done and this is our solution. So in this case, our solution is good and I know that X is equal to eight. But if I got a negative number here, this would actually not be a solution at all. So remember here, you cannot take the log of a negative number. It is not a solution if you do get a negative number that you're taking the log of. So now that we know how to solve all of these log equations and we already knew how to solve all of our exponential equations, we are good to go. And let's get some more practice. Let me know if you have questions.