Solving Logarithmic Equations

In this video, we're going to learn how to solve logarithmic equations. Let's start with the definition of a log. y equals log base a of x is equivalent to x equals a to the y and a has to be greater than zero and a cannot be equal to one for this definition to be true. to be true. We also know if log base a of m is equal to log base a of n, then m is equal to n. Now this could be that m and n are numbers, but more than likely one or both of them is going to be an expression involving a variable. Well here's the catch. m, n, and a are all positive. And also, a cannot be equal to 1. So the expression for m and n, whether it's a variable or a number, has to be positive. So in other words, when we get an answer, we must go back and check to make sure that when substituted in and simplified for the expression m and also for the expression n, that we get a positive number. If it's negative then that's considered an extraneous solution and we'll see some of those as part of our problems. We are going to start by solving these two logarithmic equations. Each of these two is a little bit different. The first one has a single logarithmic expression equal to a constant. When you have that you are going to set or convert this over to exponential form. That means you'll convert it from this logarithmic expression or equation sorry to become two to the fourth is equal to 5x, and that's using the definition of a logarithm. We'll then divide by 5 to both sides, and so what we get is x is equal to 16 fifths. Now what we need to do is check by putting this back into the equation, and really what I'm concerned with more than actually evaluating log base 2 of 5 times 16 fifths, what I'm really concerned about is what is 5 times 16 fifths. In other words, Is that going to be a negative number or a or zero? So if I simplify that expression, I get positive 16. And I can continue by checking to see that log base 2 of 16 is in fact equal to 4. But what I wanted to really do right here is verify that 5 times whatever our answer is, is positive. And it was. So therefore, 16 fifths is an accurate solution. And since log base 2 of 16 is equal to 4, we know it's correct. Now, let's go through the second one, which is a little different. The second one now has a log expression equal to another log expression without a constant, and that's very important. So here we had a constant of 4, and number 2, we don't have a constant. First thing I'm going to do is notice that I have coefficients of 2 and 3 for each of the log terms, and what I should do is use the properties of logarithms to convert those into exponents. So log... Base 5 of x squared is equal to log base 5 of 4 to the third. Now what I can do is, because of that second relationship that I just defined, since these are both logs on each side, there is no constant, and they are the same base, by the way, we can then set the expressions that we're taking the logarithms of. In other words, what we had before I said is m and n, set those equal to each other. So we know x to the second is equal to 4 to the third, which, by the way, is 64. So then we can square root both sides. This is a quadratic equation. So we'll get x is equal to plus or minus 8. So we'll get two answers. Now, we must check this. So if I put these back in, I should be getting a positive expression before you take the log. And we can see that 8 would work. So log base 5 of 8 is perfectly fine, and you can evaluate that with your calculator. But... 2 times log base 5 of negative 8 is not okay. So that is actually not an actual solution to the original equation. So because of that, negative 8 is considered an extraneous solution. So x equals positive 8 is the only solution, and x equals negative 8 is called an extraneous solution. It's an apparent solution based on our solution process, but not an actual solution. Positive 8 is the only solution to this equation. These next two problems are very similar to the first problem in the sense that what we have is a logarithmic expression on one side and a constant on the other side. Now the difference is in the first problem we had just one log term. In this problem and with number four also, we have two log terms. So what we need to do first is take our properties and figure out a way to condense these down to a single logarithmic term. So we have two logarithmic terms. They're separated by addition, and a property of logarithms says we can condense this down to one log term, and the addition tells us what we can do is take the product, in other words, multiply the two expressions we're taking the logarithm of. So x times x plus 15, and this is still equal to 2. Now from here, what we can do is go ahead and solve just like we did number 1. Now remember, this is base 10. It's not written there, so it's assumed, so you need to remember that. And what we get is x times x plus 15 is equal to 10 squared. Now, from here, what we have is a quadratic equation, which we must solve using our strategies to solving quadratic equations, which means we need to either factor or complete the squares, or we can use the quadratic formula. But regardless of which method you choose, the first thing we need to do is distribute the x. So we have x squared plus 15x equals 100, and we must get this equal to 0, so we'll have x squared plus 15x minus 100 equals 0. Now I choose to factor this quadratic, so when I do that, I get x plus 20 and x minus 5 as my two factors, which result in two solutions of negative 20 and positive 5. Again, I want to go back and check this. So if I were to substitute 5 and negative 20 back into my original equation, it should satisfy. So I'm not taking the logarithm of a negative number or 0. If I check 5, 5 plus 15 is 20, which is perfectly fine, and logarithm of 5 is fine also. I now need to check negative 20, which I can pretty much see that it fails on both of my logarithms. Now it doesn't need to fail both of them, it just needs to fail one of them. So even if one of those... happens to result in a negative number or zero that you're taking the logarithm of, then it's an extraneous solution. So your answer is x equals 5, where negative 20 is considered an extraneous solution. The last problem is very similar to number 3. So in this case, we're going to again take the addition and condense this to a single log by finding, we're taking the product of our two binomials probably the main the biggest difference between these two is the fact that my base is no longer 10 and also the fact that instead of a monomial a single term in one of my one of my factors I now have a binomial times a binomial but otherwise it's almost identical so I'm going to go ahead and I choose let's go ahead and multiply this out first so we'll have x squared plus 15x plus 56 and this is equal to 1. Now I'm going to change this into exponential form which is x squared plus 15x plus 56 is equal to 2 to the first. I need to move the 2 to the first which is 2 over to the left side so I have x squared plus 15x plus 54 is equal to 0. Again, I choose to factor, but if you can't factor, if you cannot think of the factors, because remember, not all quadratics are actually factorable, then what you can do is use the quadratic formula. But in this case, what I get is x plus 9 and x plus 6 when I factor this, which results in two solutions of negative 9 and negative 6. Now, don't just assume because of the fact that both of them are negative that both of them do not satisfy the original equation. Again, the idea is not that because the solution is negative, it fails. It's that when you simplify what you're taking the logarithm of, it needs to fail. So if I take negative 6 plus 7, if this equals a positive number, which it does, and negative 6 plus 8, if that equals a positive number, which it does, then that solution is actually fine. However, if I do this and instead of negative 6, I have my solution of negative 9, then what you see is that you get negative 2. And you get negative 1 as the values of the expression which do fail. So this is considered an extraneous solution. And x equals negative 6 is the only actual solution to the equation. Love all driver!

to be true. We also know if log base a of m is equal to log base a of n, then m is equal to n. Now this could be that m and n are numbers, but more than likely one or both of them is going to be an expression involving a variable. Well here's the catch. m, n, and a are all positive.

And also, a cannot be equal to 1. So the expression for m and n, whether it's a variable or a number, has to be positive. So in other words, when we get an answer, we must go back and check to make sure that when substituted in and simplified for the expression m and also for the expression n, that we get a positive number. If it's negative then that's considered an extraneous solution and we'll see some of those as part of our problems. We are going to start by solving these two logarithmic equations. Each of these two is a little bit different.

The first one has a single logarithmic expression equal to a constant. When you have that you are going to set or convert this over to exponential form. That means you'll convert it from this logarithmic expression or equation sorry to become two to the fourth is equal to 5x, and that's using the definition of a logarithm. We'll then divide by 5 to both sides, and so what we get is x is equal to 16 fifths.

Now what we need to do is check by putting this back into the equation, and really what I'm concerned with more than actually evaluating log base 2 of 5 times 16 fifths, what I'm really concerned about is what is 5 times 16 fifths. In other words, Is that going to be a negative number or a or zero? So if I simplify that expression, I get positive 16. And I can continue by checking to see that log base 2 of 16 is in fact equal to 4. But what I wanted to really do right here is verify that 5 times whatever our answer is, is positive. And it was.

So therefore, 16 fifths is an accurate solution. And since log base 2 of 16 is equal to 4, we know it's correct. Now, let's go through the second one, which is a little different.

The second one now has a log expression equal to another log expression without a constant, and that's very important. So here we had a constant of 4, and number 2, we don't have a constant. First thing I'm going to do is notice that I have coefficients of 2 and 3 for each of the log terms, and what I should do is use the properties of logarithms to convert those into exponents.

So log... Base 5 of x squared is equal to log base 5 of 4 to the third. Now what I can do is, because of that second relationship that I just defined, since these are both logs on each side, there is no constant, and they are the same base, by the way, we can then set the expressions that we're taking the logarithms of.

In other words, what we had before I said is m and n, set those equal to each other. So we know x to the second is equal to 4 to the third, which, by the way, is 64. So then we can square root both sides. This is a quadratic equation. So we'll get x is equal to plus or minus 8. So we'll get two answers. Now, we must check this.

So if I put these back in, I should be getting a positive expression before you take the log. And we can see that 8 would work. So log base 5 of 8 is perfectly fine, and you can evaluate that with your calculator.

But... 2 times log base 5 of negative 8 is not okay. So that is actually not an actual solution to the original equation.

So because of that, negative 8 is considered an extraneous solution. So x equals positive 8 is the only solution, and x equals negative 8 is called an extraneous solution. It's an apparent solution based on our solution process, but not an actual solution.

Positive 8 is the only solution to this equation. These next two problems are very similar to the first problem in the sense that what we have is a logarithmic expression on one side and a constant on the other side. Now the difference is in the first problem we had just one log term.

In this problem and with number four also, we have two log terms. So what we need to do first is take our properties and figure out a way to condense these down to a single logarithmic term. So we have two logarithmic terms.

They're separated by addition, and a property of logarithms says we can condense this down to one log term, and the addition tells us what we can do is take the product, in other words, multiply the two expressions we're taking the logarithm of. So x times x plus 15, and this is still equal to 2. Now from here, what we can do is go ahead and solve just like we did number 1. Now remember, this is base 10. It's not written there, so it's assumed, so you need to remember that. And what we get is x times x plus 15 is equal to 10 squared.

Now, from here, what we have is a quadratic equation, which we must solve using our strategies to solving quadratic equations, which means we need to either factor or complete the squares, or we can use the quadratic formula. But regardless of which method you choose, the first thing we need to do is distribute the x. So we have x squared plus 15x equals 100, and we must get this equal to 0, so we'll have x squared plus 15x minus 100 equals 0. Now I choose to factor this quadratic, so when I do that, I get x plus 20 and x minus 5 as my two factors, which result in two solutions of negative 20 and positive 5. Again, I want to go back and check this. So if I were to substitute 5 and negative 20 back into my original equation, it should satisfy. So I'm not taking the logarithm of a negative number or 0. If I check 5, 5 plus 15 is 20, which is perfectly fine, and logarithm of 5 is fine also.

I now need to check negative 20, which I can pretty much see that it fails on both of my logarithms. Now it doesn't need to fail both of them, it just needs to fail one of them. So even if one of those... happens to result in a negative number or zero that you're taking the logarithm of, then it's an extraneous solution.

So your answer is x equals 5, where negative 20 is considered an extraneous solution. The last problem is very similar to number 3. So in this case, we're going to again take the addition and condense this to a single log by finding, we're taking the product of our two binomials probably the main the biggest difference between these two is the fact that my base is no longer 10 and also the fact that instead of a monomial a single term in one of my one of my factors I now have a binomial times a binomial but otherwise it's almost identical so I'm going to go ahead and I choose let's go ahead and multiply this out first so we'll have x squared plus 15x plus 56 and this is equal to 1. Now I'm going to change this into exponential form which is x squared plus 15x plus 56 is equal to 2 to the first. I need to move the 2 to the first which is 2 over to the left side so I have x squared plus 15x plus 54 is equal to 0. Again, I choose to factor, but if you can't factor, if you cannot think of the factors, because remember, not all quadratics are actually factorable, then what you can do is use the quadratic formula.

But in this case, what I get is x plus 9 and x plus 6 when I factor this, which results in two solutions of negative 9 and negative 6. Now, don't just assume because of the fact that both of them are negative that both of them do not satisfy the original equation. Again, the idea is not that because the solution is negative, it fails. It's that when you simplify what you're taking the logarithm of, it needs to fail. So if I take negative 6 plus 7, if this equals a positive number, which it does, and negative 6 plus 8, if that equals a positive number, which it does, then that solution is actually fine. However, if I do this and instead of negative 6, I have my solution of negative 9, then what you see is that you get negative 2. And you get negative 1 as the values of the expression which do fail.

So this is considered an extraneous solution. And x equals negative 6 is the only actual solution to the equation. Love all driver!

Transcript for:Solving Logarithmic Equations

Transcript for:
Solving Logarithmic Equations