 In this video we are going to discuss how to solve logarithmic equations. So the most basic logarithmic equation that we can solve is an example like this one that's shown. And what you have to remember when solving a logarithmic equation is that you can always rewrite a logarithm as an exponential equation. Just look at our definition for a logarithm and you can easily see that. So in this case what happens if we rewrite it, remember is we swing the base to the other side and a logarithm is equal to an exponent. So this expression becomes 6 to the second equals what was inside of the logarithm, 2x plus 4. So the base stays the same in the exponential as it was in the logarithm and then what the logarithm was equal to becomes the exponent. Here to solve this 6 squared is 36 and so if we simply subtract 4 from each side of the expression we get 2x equals 32 divided by 2x equals 16. Always check your answer for a logarithm because remember that our domain for a logarithm is that this inside portion of a logarithm can never be negative. Here if we plug in 16 it still is positive so it will work. But we will get some cases where it will give us a solution that does not work. Now with this example we're going to solve it in the same way but we first have to combine these two logarithms. So you cannot rewrite a logarithm as an exponential until all of the logarithms on one side are written as one expression. So this is where our properties of exponents come in. Remember that when you add two logarithms together that have the same base which these do you can multiply the inside of each logarithm. So this becomes log base 2 of x over 2 times 2x is equal to 4. And if we simplify this expression inside x over 2 multiplied by 2x just becomes x squared. So we end up with log base 2 of x squared equals 4. And now that the logarithms are combined into 1 now we can swing the base to the other side. That still stays the base and the 4 becomes the exponent. So we end up with 2 to the 4th equals x squared. 2 to the 4th is 16 and so to solve this for x we can just take the square root of each side. Now when you do take the square root it should be a positive or a negative value but when we check our answers in each of these if we plug in a positive 4 that just gives us the log base 2 of 8 but if we plug in a negative 4 it gives us a log base 2 of negative 8. And remember this part inside of the logarithm can never be negative. So our only solution here is a positive 4 not the negative 4. Now looking at this third example it's a little bit different because everything here is inside of a logarithm. Unlike before we always had just a number that was by itself. So you still have the option of moving everything to one side and then rewriting it as a logarithm. But what I prefer is I prefer to use what's called the uniqueness property. So if we can combine everything on each side into one logarithm then we can cancel out the logarithms using the uniqueness property. In this case I'm going to combine this natural log of 2x and natural log of 3x using multiplication because they're added. And then on the right side I just have one expression so I don't need to combine it at all. And so what happens is once everything is inside of a logarithm as long as the bases are the same which in this case they are because it's just the unwritten E since it's the common log they can cancel each other out. That's the uniqueness property. So those cancel and on the right when I multiply 2x times 3x I get 6x squared. I guess that was on the left sorry and the right I just have 6. So solving for x I get x squared equals 1. If I take the square root x will equal a positive or a negative 1. But remember when we check up here with our x's plugging in that negative would again make this inside part negative. So I am just going to take the positive solution x equals 1. Let's look at just one final example here. We have the log base 7 of 4x minus the log base 7 of x plus 3 equals the log base 7 of x. So again what I notice is everything is inside of a log so I'm going to just combine each side into one logarithm and then cancel the logarithms out. On the left side subtraction means I'm going to have to divide. But I'm keeping the base the same and on the right side it's already written as 1. So I don't need to change that at all. Once they are rewritten as 1 I cancel the logarithms out and I just have the expression 4x over x plus 3 is equal to x. Now this is a rational equation to solve. And what I would do is I would multiply each side by the denominator x plus 3 because that will cancel out the denominator on the left. I now have 4x equals and I'm going to distribute this x. x squared plus 3x. Now this is a quadratic equation and we could solve it by completing the square but I think what I'm going to do is I'm going to use the program, the quadratic formula and just use the program on my calculator. So I wrote the whole thing out here. So we have an A of 1, a B of negative 1, and a C of 0. And if you put it into the quadratic formula you'll get two solutions. x equals 1 or x equals 0. And what you just have to make sure to do is check it in the original equation. So anything inside of a logarithm has to be positive. It can't even be equal to 0. So when we plug a 1 in we get 4, 4, and 1. So that one works. When we plug a 0 in we get 0. Automatically because of this giving us 0 we know that that is not a solution. So just x equals 1 would be the solution in this case.