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Hi, my name is Rebecca Muller. Now that have learned how to change from exponential to logarithmic form and back again, and you have some experience working with properties of logarithms, it's time to look at solving exponential and logarithmic equations. During this session, we'll start with solving exponential equations. Here are some of the methods we're going to use. We'll use the one-to-one property. We'll look at using logarithms. And we'll use substitution in equations of the quadratic type. Then we'll go to solving logarithmic equations. And we'll use exponential form and the one-to-one property. And then we'll use other properties of logarithms in order to solve logarithmic equations and exponential equations as well. So let's look at an example of how we're going to solve exponential equations. And we're going to use as our example 3 to the x equals 9. Now, that was pretty easy to solve. You probably know the answer right away. 3 squared equals 9. But let's just notice that what you're doing is you're thinking of the fact that 9 is the same as 3 raised to the second power. So at this point, because we have a one-to-one property for exponential functions, we know that the exponents have to be the same. That is, once I can get the base to be the same on both sides of the equation, I can equate the exponents, and say that x has to equal the value of 2. And that leads us to the following property. For b greater than 0, b not equal to 1, and x, y equal to real numbers, b to the x equals b to the y if and only if x equals y. So in other words, if I can get the base to be the same on both sides of the equation, then I can equate the exponent. So this was what we did in the expression that we see here. So now let's look at a second example where we're going to use this one-to-one property. Here's the equation I want to consider. We're going to have 25 raised to the x minus 1 equals 125 raised to the 2x minus 3. Now, when we look at this, it's obviously an exponential equation because our variables are in the exponents. We can look at our bases. We have 25 is the base on the left and 125 is the base on the right. Well, if you're familiar with your exponent powers, you'll recognize these as both being powers of 5. So I can rewrite 25 as 5 squared. That's inside parentheses raised to the x minus 1. I can rewrite 125 as 5 cubed. And that's in parentheses raised to the 2x minus 3. So what do we do with exponent powers when they look like this? We multiply them. So this first part becomes 5 to the 2 times x minus 1 power equals 5 to the 3 times 2x minus 3 power. Now, at this step, it's very important that you recognize that you have to use parentheses here, because I'm going to have to do distributive property in order to evaluate this. What we can now note is that I have the same base on both sides of the equation. And therefore, I can use the one-to-one property that allows me to equate the exponents. So that means that 2 times x minus 1 will equal 3 times 2x minus 3. Let's continue by using the distributive property. So 2x minus 2 equals 6x minus 9. We can subtract 6x from both sides of the equation. That's going to give us a negative 4x. And at the same time, I'm going to add 2 to both sides of the equation. So negative 9 plus 2 is negative 7. And then dividing by negative 4 on both sides of the equation gives us 7/4. Now, at this point, let's go ahead and check our result to see if we are coming up with something that is going to make sense to you. I'm going to do the check by substituting in to each side of the equation individually. So I'm going to begin with 25 raised to the-- this is going to be the 7/4 power, minus 1. Well, in order to combine these, I need a common denominator. So it's going to be 25 raised to the 7/4 minus 4/4, since 1 is equal to 4/4. And I come up with 25 raised to the 3/4 power. Now, I'm not going to try to evaluate that. I'm just going to wonder whether the right-hand side is also equivalent to this. So this is 125. It's raised to the 2 times 7/4 minus 3 power, like that. This is going to equal 125. And let's go ahead and simplify. We can take 2 times 7/4. That's going to be 14/4, which is the same as 7/2. Minus. I'm going to go ahead and rewrite the 3 in terms of halves. So 3 is the same as 6/2. And what I come up with is 125 to the 1/2 power. Now, when you look at these two, you may say, hey, I'm not sure that's the same thing. But if we do a little bit more evaluation, we can note that 25, recall, is 5 squared. So is this 5 squared to the 3/4. We multiplied the exponents together in order to give us for this result-- let me just go ahead and write it down here-- I'm going to end up with 5 to the 6/4, which is the same as 5 to the 3/2. And now this part, 125, is the same as 5 cubed. And then that's raised to the 1/2 power. Multiplying exponents gives us 5 to the 3/2 power. So we can see that these two are actually equal to each other, and therefore are solution checks. It's time for a quick quiz. Solve the exponential equation e squared divided by e to the x power equals e to the fifth. Is the correct answer. A. x equals 2 fifths. B. x equals negative 3. or C. x equals 5 halfs. Solve the equation and then choose your selection from A, B, or C. You're correct, the answer is B. x equals negative 3. Sorry the answer is B, x equals negative 3. To solve the equation we can begin on the left hand side equation by using a property of exponents. We have e squared divided by e to the x. This is going to have two expressions that have the same base. We can rewrite this as e to the 2 minus x. Since we have a division we can re-write this as a difference of the exponents. This now equals e to the fifth power. Now because of the one to one property with exponential functions we can equate exponents. So 2 minus x is equal to 5. I can subtract 2 from both sides the equation in my next step so negative x equals 3 and then dividing by negative 1 gives us x equals negative 3 as our answer. Now, as long as you're able to get the same base on both sides of an equation, then that's the way to go. We want to go ahead and equate the exponents. But there are times when that's not going to be possible. Let's look at the next example to see what we can do then. So we're going to start off with the equation 4 times 3 to the x minus 1 minus 5 equals the value 3. So first of all, I need to isolate my expression that has the x. So I'm going to go ahead and start by adding 5 to both sides of the equation. That's going to give us 4 times 3 to the x minus 1 equals the value of 8. And I still need to isolate more. I'm going to divide both sides of this equation by 4. That's going to give us 3 to the x minus 1 equals 8 divided by 4, which is 2. Now, at this point, I cannot say that 3 to the power something equals 2. I don't know that result right offhand. So I have some alternatives. One thing I can do is change this into its logarithmic form. And so note that I can say, OK, well, great. I can solve for the exponent by saying, this is log base 3 of 2 equals the exponent x minus 1. And now adding 1 to both sides of the equation, rewrite this as 1 plus log base 3 of 2 equals x. Now, that's an exact answer. If I want an approximation, I can now use the base-changing formula to give us a value here, and then substitute that into the calculator. So this is going to be 1 plus, and I'm going to choose to introduce the base e. So using the base-changing formula, I have the natural log of 2 divided by the natural log of 3 equals x. Again, this is still an exact answer until I use that calculator. Using your calculator, you should come up with a value of x which is approximately equal to 1.631. And I'm going to leave the check up to you at this point. But it turns out there's an alternative way to do this problem that you may prefer. And that is at this point right here, where I start having to use logarithms, I can also use the one-to-one property of logarithms in order to help us. And that is, I can take that step-- let me just go ahead and rewrite it here. I have 3 to the x minus 1 equals 2. And I can take the log of both sides of the equation. So again, if I decide to use natural log as my go-to, I can say that the natural log of 3 to the x minus 1 equals the natural log of 2. Now I'm able to use the properties of logarithms to help me out. I can take x minus 1 and pull it in front of the factor. And that's written as x minus 1 times the natural log of 3 equals the natural log of 2. Now, don't be concerned about the fact that the natural log of 3 looks a little bit more difficult than just a number. It's just a number. So natural log of 3 is going to be now distributed into this difference here in order to give us x times the natural log of 3 minus the natural log of 3 equals the natural log of 2. Now, notice that my term with x in it can be isolated. I'm going to add natural log of 3 to both sides of the equation. So x times the natural log of 3 equals the natural log of 2 plus the natural log of 3. Now I can solve for x by dividing both sides of the equation by the natural log of 3. So x is going to equal the natural log of 2 plus the natural log of 3 divided by the natural log of 3. Now, that may not look exactly like what we had over here, but notice that I have two terms in the numerator being divided by a single term in the denominator, so I can divide each term in the numerator by the denominator. That looks like this. x equals the natural log of 2 divided by the natural log of 3 plus the natural log of 3 divided by the natural log of 3. And so notice if you want to see that they are comparable to each other, you can do these last two steps in order to see that the natural log of 2 divided by the natural log of 3 plus 1 ends up being the same result, regardless of the method that I chose. Probably if you go this route, you could stop with the exact answer right here. There's really no reason to have to go to the last two. Again, my main reason for doing that was to point out to you that you do end up with the same solution, regardless of the method that you choose. Now, if it turns out that you're looking at an exponential equation where the variable is in two different terms and you cannot combine those together, then it could happen that you're ending up looking at an equation of the quadratic type, and so I'm going to demonstrate that to you now. Here's the example. If we have e to the 2x plus e to the x minus 2 equals 0, notice I have two terms that have the variable in them, but I'm not able to combine those two terms. They're not like terms. But what I can notice is that e to the 2x is the same as e to the x squared. And so what that allows me to do is to use a u substitution in order to solve this equation. Here's what it looks like. Let's let u equal e to the x power. Now our first term can be rewritten as u squared. Plus, my e to the x is u, and then minus 2 equals 0. Now, once I have it in this format, it looks like a quadratic equation because this is an equation of the quadratic type. I can use the process that I would use on a quadratic equation to solve my exponential one. Let's factor this. u squared can be factored as u times u. The value of 2 can be factored as 2 times 1. I need to end up with a plus 1u in the middle term. So if I add 2 u and subtract 1 u-- that's my inner and my outer terms-- I'll end up with my middle term, and just double checking my last term, I do end up with a negative 2. That means that u is equal to negative 2, or u is equal to 1. Now, remember, in a substitution problem at this point, we now need to sometimes what we call back-substitute. That is, we need to replace our u with the e to the x. So at this point, I'm going to rewrite with the two equations that we end up with. That's going to be e to the x equals negative 2 or e to the x equals 1. Let's talk about the first one. We have a positive number raised to a power. It cannot equal a negative 2. So we're not going to end up with any solutions from that first part. How about the second one? Well, if you just look at it by inspection, you may note that e to the 0 power is going to end up giving you 1, and so you come up with x equals 0. The other alternative is to change it to its logarithmic form. That would be log base e, which is our natural log, of 1 equals x. And now using the property of logarithms, we know that the log of 1 is equal to 0. So either way you look at it, you come up with x equals 0 as a solution. Let's check our work. We're going to substitute into our original equation. So we're going to have e to the power 2 times 0, which is going to be 0, plus e to the power of 0, minus 2 equals 0. e to the 0 power is 1, so 1 plus 1 minus 2 equals 0. And 2 minus 2 equals 0. And so we can see that our solution does indeed check. Now we're going to move on to logarithmic equations. We're going to begin with what are called simple log equations. These are the ones where we're able to isolate the logarithm and then change to exponential form. In order to solve. Here's our first example. Let's look at 2 times log base 3 of x minus 1 equals 0. Now, we can isolate the logarithmic expression, which is right here. And notice that if we just think about this as a variable itself, it's pretty easy to see how to do that. We can add 1 to both sides of the equation. So to start with, we have 2 times log base 3 of x equals 1. And then divide by 2. That's going to give us log base 3 of x equals 1/2. At this point, again, we have the logarithm isolated. We're now able to change this into exponential format. That's going to give us 3 to the 1/2 power equals x. And recall that the 1/2 power means that what we have is a square root of 3. Let's check that work. Go back to the original setup. We're going to have 2 multiplied times log base 3 of the square root of 3 minus 1. And let's see what we end up with. Let's make sure we end up with 0. So we have to evaluate log base 3 of the square root of 3. Remember that means we're trying to find the exponent power that we put on 3 in order to end up with the square root of 3. And that, again, is going to reiterate the idea that the square root is really the 1/2 power. So the exponent to put on 3 to end up with square root of 3 is 1/2. And now 2 times 1/2 is going to be 1, and 1 minus 1 equals 0. So our solution checks. As a second example, let's consider log base x of 25 equals 2. Well, I already have the logarithm isolated. So this is going to be in simple log form. I can change this into its exponential format by writing x squared equals 25. Now, this is a quadratic that can be solved by taking the square root of both sides of the equation. And if you recall, when we take square root of both sides of the equation, we have to get the plus or minus square root of 25, which is 5. Now, here's where it's imperative to think about the fact that logarithms have restrictions on them. Namely, we know that the base has to be a positive value. So even though this quadratic gives us two solutions, the negative solution cannot be substituted into the original format of the equation. So we have to take as our solution simply the positive value of 5. Again, checking our work, we come up with log base 5 of 25. What is the exponent we put on 5 to end up with 25? And the result is 2, so this checks also. Now, in some logarithmic equations, we're going to use the one-to-one properties of logs in order to help us. Here's an example of this. If we look at the natural log of 2x minus 5 equals the natural log of x plus 7, then we can see that because the natural log of something equals the natural log of something, we can equate the somethings. That is, we can say because of the one-to-one property that 2x minus 5 must equal x plus 7. We can now simplify this by subtracting x from both sides of the equation. That's going to give us 2x minus x, which is x. And at the same time, I'm going to go ahead and add 5 to both sides of the equation. 7 plus 5 gives us the value 12. So let's just check our work. If we look at the natural log of 2 times 12 minus 5, that's going to give us the natural log of 24 minus 5, which is the natural log of 19. On the right-hand side, we're looking at the natural log of 12 plus 7, which is also the natural log of 19. And so we can see that our solution does indeed check. Sometimes logarithmic equations require the use of properties of logarithms in order to solve them. Here's an example of this. Let's look at log base 3 of x plus log base 3 of x minus 4 equals log base 3 of 5. So on the left-hand side, we see that we have a sum of two logarithms. We can combine that into the log of the product. So this is log base 3 of x times x minus 4. On the right-hand side, we'll simply have log base 3 of 5. Now, I'm just going to go ahead and put it in an extra step here. I'm going to multiply out inside the parentheses here. That will give us log base 3 of x squared minus 4x equals log base 3 of 5. And I say it was an extra step because I could have used the one-to-one property right here, but I'm certainly going to show you that now. I have log base 3 on both sides of the equation, so by the one-to-one property, I can equate x squared minus 4x with 5. And so those two have to equal each other. Now we're looking at a quadratic equation. So we'll set it equal to 0, and factor. x squared can be factored as x times x. 5 can be factored as 5 times 1. And in order to end up with our middle term of negative 4x, I'll put the minus sign in front of the 5 to give us a negative 5x as our inner term, and the plus sign in front of the 1 to give us the plus 1x. And that works out to give us the negative 5 as the last term. And so recall when we're here, we now have that x equals 5 to make this first factor equal to 0, or we have x equals negative 1 to make the second factor equal to 0. Now, it's really important for us to remember that we're dealing with logarithmic expressions, and that logarithmic expressions have restrictions on their domains. So when we get these answers, we have to check our work. And so just starting off with the value of 5, I'm going to have log base 3 of 5 plus, this is going to be log base 3. And I'm going to go ahead and just evaluate. 5 minus 4, which is equal to 1. And I know that the log of 1 is going to equal 0, so I end up with log base 3 of 5 plus 0. And of course, that does equal the right-hand side, log base 3 of 5. So what we find out is for the solution x equals 5, this is going to work. What about the second solution, which is x equals negative 1? Well, I'll start by substituting it in. And the first thing I get is log base 3 of negative 1. Well, I don't have to go any further. This is not going to be possible, as we only can take the logarithm of positive values. So this solution has to go away. And the only solution to this equation is x equals 5. It's time for another quick quiz. Solve the equation log base 3 of 2 plus log base 3 of x equals 2. Is the answer given in A, x equals 9 halfs or in B, x equals 3 or in C, x equals 7. Solve the equation then choose your answer from A, B, or C. You're correct. The answer is A, x equals 9 halfs. Sorry the answer is A, x equals 9 halfs. Let's begin by looking at the left hand side the equation we have a sum of 2 logarithms those can be combined into a single logarithm where I'm going to take the log of the product so I multiply 2 times x this now reads log base 3 of 2x equals the value 2. Next I can change this logarithmic equation into its exponential form so my base is 3 my exponent is 2 and that has to equal 2x so 3 squared equals 2x when 9 equals 2x. To solve for x requires to be divide both sides the equation by 2 and so 9 halfs equals x is our result. Now we're going to look at a logarithmic equation, which is going to do a combination of things we've seen previously. Let's start off with the equation log base 9 of x plus 2 minus log base 9 of x equals 1/2. At the outset, what do you notice on the left-hand side of the equation? You should see that we have a difference of two logarithms. So when we have a difference of logs, what can we change it to? We can change it to the log of a quotient. Let's see what that looks like now. We're going to have log base 9 of the quotient x plus 2 divided by x equals 1/2. Well, at this point, notice that I have a simple log equation that can be changed to exponential form. So our base is 9. The exponent is 1/2. And it will equal the expression that has the variables in it. What can we do now? Well, don't forget you can always evaluate things. So let's evaluate 9 to the 1/2 power. That's simply the value 3. At this point, we can now multiply through by the denominator of x in order to give us our next step. So let's multiply by the LCD. That's going to give us 3x equals x plus 2. We'll subtract x from both sides of the equation to give us 2x equals 2, and then divide by 2 to give us x equals 1. Now, remember, we want to check our work. So we're going to take our expression x equals 1 and we're going to substitute it back into the original equation. So we're going to end up with log base 9 of 1 plus 2 minus log base 9 of 1 on the left-hand side of the equation. We can add 1 and 2. That's going to give us log base 9 of 3. And we can evaluate log base 9 of 1, because the log of 1 equals 0. And now in our next step, remember what we have to think about is, what's going to happen with log base 9 of 3? What is the exponent power we need to raise 9 to in order to end up with the value 3? That result is 1/2. And so this solution does indeed check. Now it's time for you to try a problem on your own. Here's the equation I'd like you to attempt to solve. We want to pause the video, and try it on your own, and then come back. Let me read it to you. We have 2 log base 4 of x equals log base 4 of the quantity x minus 1 plus 1. Try it and then come back. Let's see how you did. Well, the first thing I notice is that I have the plus 1 over to the right, which is just a constant. I need to get those logarithms on the same side of the equation. While I'm doing that, let's just note that this term 2 times log base 4 of x can be rewritten as the 2 with the exponent power. So we're going to rewrite that in one step. It's going to read log base 4 of x squared. And I'm going to subtract log base 4 of x minus 1. And that's going to equal the value of 1. So just looking at this step again, there are really two things I did in one. I used the log of a power and pulled that exponent into the exponent here. And then I subtracted logarithms. So now what do we do with the difference of two logs? We can rewrite that as a log of a quotient. So we'll have log base 4 of the quotient where I have x squared divided by x minus 1. This is now a simple log equation. We have the logarithm isolated equal to a number. We can change this to its exponential form. So our base is 4. Our exponent is 1. And this equals x squared divided by x minus 1. And of course, 4 to the 1 is simply 4. So at this point, I can multiply through by the denominator. 4 times x minus 1 equals x squared. Let's use distributive property and continue with the problem up here. We're going to have 4x minus 4 equals x squared. Now, I recognize this as a quadratic equation. So I can choose to set this equal to 0, and I'm just going to do that by leaving-- the terms on the left, I'm going to move to the right, and have the left-hand side equal 0. So this is going to be x squared minus 4x and then plus 4. We can factor that. And you may recognize it as a special product. This is x minus 2 quantity squared. If you used FOIL and wrote it as x minus 2 times x minus 2, that's fine. At this point, we take that factor and set it equal to 0, and that gives us the result x equals 2. Now, remember, we always want to check our work when we're dealing with logarithmic expressions. So let's go to our check now. We're going to take the value x equals 2 and substitute it into the original equation. So that's going to give us 2 times log base 4 of 2 equals maybe log base 4 of 2 minus 1 plus 1. So instead of doing each side independently, I'm just going to go ahead and write them down together. But remember, I'm really checking to see whether or not they're equivalent. On the left-hand side, we have 2 times-- and I'm going to evaluate log base 4 of 2, which means I'm looking for the exponent that I put on the base 4 to end up with 2. And that is going to be 1/2, since the square root of 4 equals 2. On the right-hand side, we'll have log base 4, 2 minus 1, which is 1, plus 1. 2 times 1/2 is 1. The log of 1 is always 0, and so 1 does indeed equal 0 plus 1, and we've checked our work. So now you've had some experience dealing with the solutions of exponential and logarithmic equations. Try some problems now on your own to see how well you do. Good luck.

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Hi, my name is Rebecca Muller. Now that have learned how to change from exponential to logarithmic form and back again, and you have some experience working with properties of logarithms, it's time to look at solving exponential and logarithmic equations. During this session, we'll start with solving exponential equations. Here are some of the methods we're going to use. We'll use the one-to-one property. We'll look at using logarithms. And we'll use substitution in equations of the quadratic type. Then we'll go to solving logarithmic equations. And we'll use exponential form and the one-to-one property. And then we'll use other properties of logarithms in order to solve logarithmic equations and exponential equations as well. So let's look at an example of how we're going to solve exponential equations. And we're going to use as our example 3 to the x equals 9. Now, that was pretty easy to solve. You probably know the answer right away. 3 squared equals 9. But let's just notice that what you're doing is you're thinking of the fact that 9 is the same as 3 raised to the second power. So at this point, because we have a one-to-one property for exponential functions, we know that the exponents have to be the same. That is, once I can get the base to be the same on both sides of the equation, I can equate the exponents, and say that x has to equal the value of 2. And that leads us to the following property. For b greater than 0, b not equal to 1, and x, y equal to real numbers, b to the x equals b to the y if and only if x equals y. So in other words, if I can get the base to be the same on both sides of the equation, then I can equate the exponent. So this was what we did in the expression that we see here. So now let's look at a second example where we're going to use this one-to-one property. Here's the equation I want to consider. We're going to have 25 raised to the x minus 1 equals 125 raised to the 2x minus 3. Now, when we look at this, it's obviously an exponential equation because our variables are in the exponents. We can look at our bases. We have 25 is the base on the left and 125 is the base on the right. Well, if you're familiar with your exponent powers, you'll recognize these as both being powers of 5. So I can rewrite 25 as 5 squared. That's inside parentheses raised to the x minus 1. I can rewrite 125 as 5 cubed. And that's in parentheses raised to the 2x minus 3. So what do we do with exponent powers when they look like this? We multiply them. So this first part becomes 5 to the 2 times x minus 1 power equals 5 to the 3 times 2x minus 3 power. Now, at this step, it's very important that you recognize that you have to use parentheses here, because I'm going to have to do distributive property in order to evaluate this. What we can now note is that I have the same base on both sides of the equation. And therefore, I can use the one-to-one property that allows me to equate the exponents. So that means that 2 times x minus 1 will equal 3 times 2x minus 3. Let's continue by using the distributive property. So 2x minus 2 equals 6x minus 9. We can subtract 6x from both sides of the equation. That's going to give us a negative 4x. And at the same time, I'm going to add 2 to both sides of the equation. So negative 9 plus 2 is negative 7. And then dividing by negative 4 on both sides of the equation gives us 7/4. Now, at this point, let's go ahead and check our result to see if we are coming up with something that is going to make sense to you. I'm going to do the check by substituting in to each side of the equation individually. So I'm going to begin with 25 raised to the-- this is going to be the 7/4 power, minus 1. Well, in order to combine these, I need a common denominator. So it's going to be 25 raised to the 7/4 minus 4/4, since 1 is equal to 4/4. And I come up with 25 raised to the 3/4 power. Now, I'm not going to try to evaluate that. I'm just going to wonder whether the right-hand side is also equivalent to this. So this is 125. It's raised to the 2 times 7/4 minus 3 power, like that. This is going to equal 125. And let's go ahead and simplify. We can take 2 times 7/4. That's going to be 14/4, which is the same as 7/2. Minus. I'm going to go ahead and rewrite the 3 in terms of halves. So 3 is the same as 6/2. And what I come up with is 125 to the 1/2 power. Now, when you look at these two, you may say, hey, I'm not sure that's the same thing. But if we do a little bit more evaluation, we can note that 25, recall, is 5 squared. So is this 5 squared to the 3/4. We multiplied the exponents together in order to give us for this result-- let me just go ahead and write it down here-- I'm going to end up with 5 to the 6/4, which is the same as 5 to the 3/2. And now this part, 125, is the same as 5 cubed. And then that's raised to the 1/2 power. Multiplying exponents gives us 5 to the 3/2 power. So we can see that these two are actually equal to each other, and therefore are solution checks. It's time for a quick quiz. Solve the exponential equation e squared divided by e to the x power equals e to the fifth. Is the correct answer. A. x equals 2 fifths. B. x equals negative 3. or C. x equals 5 halfs. Solve the equation and then choose your selection from A, B, or C. You're correct, the answer is B. x equals negative 3. Sorry the answer is B, x equals negative 3. To solve the equation we can begin on the left hand side equation by using a property of exponents. We have e squared divided by e to the x. This is going to have two expressions that have the same base. We can rewrite this as e to the 2 minus x. Since we have a division we can re-write this as a difference of the exponents. This now equals e to the fifth power. Now because of the one to one property with exponential functions we can equate exponents. So 2 minus x is equal to 5. I can subtract 2 from both sides the equation in my next step so negative x equals 3 and then dividing by negative 1 gives us x equals negative 3 as our answer. Now, as long as you're able to get the same base on both sides of an equation, then that's the way to go. We want to go ahead and equate the exponents. But there are times when that's not going to be possible. Let's look at the next example to see what we can do then. So we're going to start off with the equation 4 times 3 to the x minus 1 minus 5 equals the value 3. So first of all, I need to isolate my expression that has the x. So I'm going to go ahead and start by adding 5 to both sides of the equation. That's going to give us 4 times 3 to the x minus 1 equals the value of 8. And I still need to isolate more. I'm going to divide both sides of this equation by 4. That's going to give us 3 to the x minus 1 equals 8 divided by 4, which is 2. Now, at this point, I cannot say that 3 to the power something equals 2. I don't know that result right offhand. So I have some alternatives. One thing I can do is change this into its logarithmic form. And so note that I can say, OK, well, great. I can solve for the exponent by saying, this is log base 3 of 2 equals the exponent x minus 1. And now adding 1 to both sides of the equation, rewrite this as 1 plus log base 3 of 2 equals x. Now, that's an exact answer. If I want an approximation, I can now use the base-changing formula to give us a value here, and then substitute that into the calculator. So this is going to be 1 plus, and I'm going to choose to introduce the base e. So using the base-changing formula, I have the natural log of 2 divided by the natural log of 3 equals x. Again, this is still an exact answer until I use that calculator. Using your calculator, you should come up with a value of x which is approximately equal to 1.631. And I'm going to leave the check up to you at this point. But it turns out there's an alternative way to do this problem that you may prefer. And that is at this point right here, where I start having to use logarithms, I can also use the one-to-one property of logarithms in order to help us. And that is, I can take that step-- let me just go ahead and rewrite it here. I have 3 to the x minus 1 equals 2. And I can take the log of both sides of the equation. So again, if I decide to use natural log as my go-to, I can say that the natural log of 3 to the x minus 1 equals the natural log of 2. Now I'm able to use the properties of logarithms to help me out. I can take x minus 1 and pull it in front of the factor. And that's written as x minus 1 times the natural log of 3 equals the natural log of 2. Now, don't be concerned about the fact that the natural log of 3 looks a little bit more difficult than just a number. It's just a number. So natural log of 3 is going to be now distributed into this difference here in order to give us x times the natural log of 3 minus the natural log of 3 equals the natural log of 2. Now, notice that my term with x in it can be isolated. I'm going to add natural log of 3 to both sides of the equation. So x times the natural log of 3 equals the natural log of 2 plus the natural log of 3. Now I can solve for x by dividing both sides of the equation by the natural log of 3. So x is going to equal the natural log of 2 plus the natural log of 3 divided by the natural log of 3. Now, that may not look exactly like what we had over here, but notice that I have two terms in the numerator being divided by a single term in the denominator, so I can divide each term in the numerator by the denominator. That looks like this. x equals the natural log of 2 divided by the natural log of 3 plus the natural log of 3 divided by the natural log of 3. And so notice if you want to see that they are comparable to each other, you can do these last two steps in order to see that the natural log of 2 divided by the natural log of 3 plus 1 ends up being the same result, regardless of the method that I chose. Probably if you go this route, you could stop with the exact answer right here. There's really no reason to have to go to the last two. Again, my main reason for doing that was to point out to you that you do end up with the same solution, regardless of the method that you choose. Now, if it turns out that you're looking at an exponential equation where the variable is in two different terms and you cannot combine those together, then it could happen that you're ending up looking at an equation of the quadratic type, and so I'm going to demonstrate that to you now. Here's the example. If we have e to the 2x plus e to the x minus 2 equals 0, notice I have two terms that have the variable in them, but I'm not able to combine those two terms. They're not like terms. But what I can notice is that e to the 2x is the same as e to the x squared. And so what that allows me to do is to use a u substitution in order to solve this equation. Here's what it looks like. Let's let u equal e to the x power. Now our first term can be rewritten as u squared. Plus, my e to the x is u, and then minus 2 equals 0. Now, once I have it in this format, it looks like a quadratic equation because this is an equation of the quadratic type. I can use the process that I would use on a quadratic equation to solve my exponential one. Let's factor this. u squared can be factored as u times u. The value of 2 can be factored as 2 times 1. I need to end up with a plus 1u in the middle term. So if I add 2 u and subtract 1 u-- that's my inner and my outer terms-- I'll end up with my middle term, and just double checking my last term, I do end up with a negative 2. That means that u is equal to negative 2, or u is equal to 1. Now, remember, in a substitution problem at this point, we now need to sometimes what we call back-substitute. That is, we need to replace our u with the e to the x. So at this point, I'm going to rewrite with the two equations that we end up with. That's going to be e to the x equals negative 2 or e to the x equals 1. Let's talk about the first one. We have a positive number raised to a power. It cannot equal a negative 2. So we're not going to end up with any solutions from that first part. How about the second one? Well, if you just look at it by inspection, you may note that e to the 0 power is going to end up giving you 1, and so you come up with x equals 0. The other alternative is to change it to its logarithmic form. That would be log base e, which is our natural log, of 1 equals x. And now using the property of logarithms, we know that the log of 1 is equal to 0. So either way you look at it, you come up with x equals 0 as a solution. Let's check our work. We're going to substitute into our original equation. So we're going to have e to the power 2 times 0, which is going to be 0, plus e to the power of 0, minus 2 equals 0. e to the 0 power is 1, so 1 plus 1 minus 2 equals 0. And 2 minus 2 equals 0. And so we can see that our solution does indeed check. Now we're going to move on to logarithmic equations. We're going to begin with what are called simple log equations. These are the ones where we're able to isolate the logarithm and then change to exponential form. In order to solve. Here's our first example. Let's look at 2 times log base 3 of x minus 1 equals 0. Now, we can isolate the logarithmic expression, which is right here. And notice that if we just think about this as a variable itself, it's pretty easy to see how to do that. We can add 1 to both sides of the equation. So to start with, we have 2 times log base 3 of x equals 1. And then divide by 2. That's going to give us log base 3 of x equals 1/2. At this point, again, we have the logarithm isolated. We're now able to change this into exponential format. That's going to give us 3 to the 1/2 power equals x. And recall that the 1/2 power means that what we have is a square root of 3. Let's check that work. Go back to the original setup. We're going to have 2 multiplied times log base 3 of the square root of 3 minus 1. And let's see what we end up with. Let's make sure we end up with 0. So we have to evaluate log base 3 of the square root of 3. Remember that means we're trying to find the exponent power that we put on 3 in order to end up with the square root of 3. And that, again, is going to reiterate the idea that the square root is really the 1/2 power. So the exponent to put on 3 to end up with square root of 3 is 1/2. And now 2 times 1/2 is going to be 1, and 1 minus 1 equals 0. So our solution checks. As a second example, let's consider log base x of 25 equals 2. Well, I already have the logarithm isolated. So this is going to be in simple log form. I can change this into its exponential format by writing x squared equals 25. Now, this is a quadratic that can be solved by taking the square root of both sides of the equation. And if you recall, when we take square root of both sides of the equation, we have to get the plus or minus square root of 25, which is 5. Now, here's where it's imperative to think about the fact that logarithms have restrictions on them. Namely, we know that the base has to be a positive value. So even though this quadratic gives us two solutions, the negative solution cannot be substituted into the original format of the equation. So we have to take as our solution simply the positive value of 5. Again, checking our work, we come up with log base 5 of 25. What is the exponent we put on 5 to end up with 25? And the result is 2, so this checks also. Now, in some logarithmic equations, we're going to use the one-to-one properties of logs in order to help us. Here's an example of this. If we look at the natural log of 2x minus 5 equals the natural log of x plus 7, then we can see that because the natural log of something equals the natural log of something, we can equate the somethings. That is, we can say because of the one-to-one property that 2x minus 5 must equal x plus 7. We can now simplify this by subtracting x from both sides of the equation. That's going to give us 2x minus x, which is x. And at the same time, I'm going to go ahead and add 5 to both sides of the equation. 7 plus 5 gives us the value 12. So let's just check our work. If we look at the natural log of 2 times 12 minus 5, that's going to give us the natural log of 24 minus 5, which is the natural log of 19. On the right-hand side, we're looking at the natural log of 12 plus 7, which is also the natural log of 19. And so we can see that our solution does indeed check. Sometimes logarithmic equations require the use of properties of logarithms in order to solve them. Here's an example of this. Let's look at log base 3 of x plus log base 3 of x minus 4 equals log base 3 of 5. So on the left-hand side, we see that we have a sum of two logarithms. We can combine that into the log of the product. So this is log base 3 of x times x minus 4. On the right-hand side, we'll simply have log base 3 of 5. Now, I'm just going to go ahead and put it in an extra step here. I'm going to multiply out inside the parentheses here. That will give us log base 3 of x squared minus 4x equals log base 3 of 5. And I say it was an extra step because I could have used the one-to-one property right here, but I'm certainly going to show you that now. I have log base 3 on both sides of the equation, so by the one-to-one property, I can equate x squared minus 4x with 5. And so those two have to equal each other. Now we're looking at a quadratic equation. So we'll set it equal to 0, and factor. x squared can be factored as x times x. 5 can be factored as 5 times 1. And in order to end up with our middle term of negative 4x, I'll put the minus sign in front of the 5 to give us a negative 5x as our inner term, and the plus sign in front of the 1 to give us the plus 1x. And that works out to give us the negative 5 as the last term. And so recall when we're here, we now have that x equals 5 to make this first factor equal to 0, or we have x equals negative 1 to make the second factor equal to 0. Now, it's really important for us to remember that we're dealing with logarithmic expressions, and that logarithmic expressions have restrictions on their domains. So when we get these answers, we have to check our work. And so just starting off with the value of 5, I'm going to have log base 3 of 5 plus, this is going to be log base 3. And I'm going to go ahead and just evaluate. 5 minus 4, which is equal to 1. And I know that the log of 1 is going to equal 0, so I end up with log base 3 of 5 plus 0. And of course, that does equal the right-hand side, log base 3 of 5. So what we find out is for the solution x equals 5, this is going to work. What about the second solution, which is x equals negative 1? Well, I'll start by substituting it in. And the first thing I get is log base 3 of negative 1. Well, I don't have to go any further. This is not going to be possible, as we only can take the logarithm of positive values. So this solution has to go away. And the only solution to this equation is x equals 5. It's time for another quick quiz. Solve the equation log base 3 of 2 plus log base 3 of x equals 2. Is the answer given in A, x equals 9 halfs or in B, x equals 3 or in C, x equals 7. Solve the equation then choose your answer from A, B, or C. You're correct. The answer is A, x equals 9 halfs. Sorry the answer is A, x equals 9 halfs. Let's begin by looking at the left hand side the equation we have a sum of 2 logarithms those can be combined into a single logarithm where I'm going to take the log of the product so I multiply 2 times x this now reads log base 3 of 2x equals the value 2. Next I can change this logarithmic equation into its exponential form so my base is 3 my exponent is 2 and that has to equal 2x so 3 squared equals 2x when 9 equals 2x. To solve for x requires to be divide both sides the equation by 2 and so 9 halfs equals x is our result. Now we're going to look at a logarithmic equation, which is going to do a combination of things we've seen previously. Let's start off with the equation log base 9 of x plus 2 minus log base 9 of x equals 1/2. At the outset, what do you notice on the left-hand side of the equation? You should see that we have a difference of two logarithms. So when we have a difference of logs, what can we change it to? We can change it to the log of a quotient. Let's see what that looks like now. We're going to have log base 9 of the quotient x plus 2 divided by x equals 1/2. Well, at this point, notice that I have a simple log equation that can be changed to exponential form. So our base is 9. The exponent is 1/2. And it will equal the expression that has the variables in it. What can we do now? Well, don't forget you can always evaluate things. So let's evaluate 9 to the 1/2 power. That's simply the value 3. At this point, we can now multiply through by the denominator of x in order to give us our next step. So let's multiply by the LCD. That's going to give us 3x equals x plus 2. We'll subtract x from both sides of the equation to give us 2x equals 2, and then divide by 2 to give us x equals 1. Now, remember, we want to check our work. So we're going to take our expression x equals 1 and we're going to substitute it back into the original equation. So we're going to end up with log base 9 of 1 plus 2 minus log base 9 of 1 on the left-hand side of the equation. We can add 1 and 2. That's going to give us log base 9 of 3. And we can evaluate log base 9 of 1, because the log of 1 equals 0. And now in our next step, remember what we have to think about is, what's going to happen with log base 9 of 3? What is the exponent power we need to raise 9 to in order to end up with the value 3? That result is 1/2. And so this solution does indeed check. Now it's time for you to try a problem on your own. Here's the equation I'd like you to attempt to solve. We want to pause the video, and try it on your own, and then come back. Let me read it to you. We have 2 log base 4 of x equals log base 4 of the quantity x minus 1 plus 1. Try it and then come back. Let's see how you did. Well, the first thing I notice is that I have the plus 1 over to the right, which is just a constant. I need to get those logarithms on the same side of the equation. While I'm doing that, let's just note that this term 2 times log base 4 of x can be rewritten as the 2 with the exponent power. So we're going to rewrite that in one step. It's going to read log base 4 of x squared. And I'm going to subtract log base 4 of x minus 1. And that's going to equal the value of 1. So just looking at this step again, there are really two things I did in one. I used the log of a power and pulled that exponent into the exponent here. And then I subtracted logarithms. So now what do we do with the difference of two logs? We can rewrite that as a log of a quotient. So we'll have log base 4 of the quotient where I have x squared divided by x minus 1. This is now a simple log equation. We have the logarithm isolated equal to a number. We can change this to its exponential form. So our base is 4. Our exponent is 1. And this equals x squared divided by x minus 1. And of course, 4 to the 1 is simply 4. So at this point, I can multiply through by the denominator. 4 times x minus 1 equals x squared. Let's use distributive property and continue with the problem up here. We're going to have 4x minus 4 equals x squared. Now, I recognize this as a quadratic equation. So I can choose to set this equal to 0, and I'm just going to do that by leaving-- the terms on the left, I'm going to move to the right, and have the left-hand side equal 0. So this is going to be x squared minus 4x and then plus 4. We can factor that. And you may recognize it as a special product. This is x minus 2 quantity squared. If you used FOIL and wrote it as x minus 2 times x minus 2, that's fine. At this point, we take that factor and set it equal to 0, and that gives us the result x equals 2. Now, remember, we always want to check our work when we're dealing with logarithmic expressions. So let's go to our check now. We're going to take the value x equals 2 and substitute it into the original equation. So that's going to give us 2 times log base 4 of 2 equals maybe log base 4 of 2 minus 1 plus 1. So instead of doing each side independently, I'm just going to go ahead and write them down together. But remember, I'm really checking to see whether or not they're equivalent. On the left-hand side, we have 2 times-- and I'm going to evaluate log base 4 of 2, which means I'm looking for the exponent that I put on the base 4 to end up with 2. And that is going to be 1/2, since the square root of 4 equals 2. On the right-hand side, we'll have log base 4, 2 minus 1, which is 1, plus 1. 2 times 1/2 is 1. The log of 1 is always 0, and so 1 does indeed equal 0 plus 1, and we've checked our work. So now you've had some experience dealing with the solutions of exponential and logarithmic equations. Try some problems now on your own to see how well you do. Good luck.