 This video is part three of solving equations of logarithmic and exponential type. We've studied solving exponential functions by changing the base by using the power property and by graphing. Now we're going to take a look at the log problems and to solve logs most of the time it's really easy to convert it to an exponential first and then use all the exponential rules that we know about solving equations. If we do have a common log or a natural log we could go straight to our calculator to be able to solve those and graph them or any other way that we wanted to solve them. But most of them will have some other base, so we'll need to convert them to exponentials. So to do that we need to isolate the logarithm just like we were isolating the base and then move the other pieces away so that we get the log with the base by itself and then we convert it and change it to an exponential. And when we're converting remember how we hopped across the pond when we were converting exponentials into logs? We do the same thing here. We know that an exponential starts with a base, so we start at the base, hop across the pond and get b to the y and then that's equal to and when we hop up across the pond again we find out that it's equal to x. So remember that logs are exponents. Logs are equal to exponents. And that's why we knew that the y was going to be the exponent on b because a log was equal to the exponent. So let's convert and see what we can do. So if we hop across the equal sign, we start with our base because it's an exponential equation we're trying to go to. We hop across to get our exponent and then we hop back to find that it's equal to x. And when you have a problem like this, 2 to the 3rd, well we know what that is and if you didn't you could go to your calculator, but 2 to the 3rd, 2 times 2 times 2 is 8. So x is equal to 8. In the second equation sometimes your base is your unknown. So we still do the same thing. We start with our base, hop across the equal sign to grab the exponent and then hop back to find out what it's equal to. And when your base is your unknown, well then that's when we want to take the root of both sides. So x then would be equal to 7. And remember it's not negative 7 because again the base has to be greater than 0. We talked about that earlier. So it's only going to be positive 7. And then in this last case we have ln x equals 7, which if I convert it, remember natural log means that it's going to be base e to the 7 is equal to x. And that I have to take to my calculator because I don't happen to know base e in my head very easily. But second ln will give me e and then I put my 7 in for the exponent and I find out that e to the 7th is approximately 1096.63, which is approximately equal to x. So there we have the what happens when we get a more involved looking log problem. Again we peel the layers. This 5 is right next to the log. This 4 has a plus in the middle so we have to move it far the same way first. Outside layer would be plus 4 of that onion. So we have 5 log of 2x is equal to subtracting 4 from both sides. We'd have negative 1. And if I divide by 5, because remember we want to get this log all by itself, so we have to divide the 5 off, now we have log of 2x is equal to negative 1 5th and we can convert it. So remember this is a supposed or assumed, not supposed, assumed 10 when it's just log. And then we hop across the pond to get the exponent of negative 1 5th equal to hop back across the pond equal to 2x. And if we divide everything by 2 we can finally get x by itself. So 10 to the negative 1 5th divided by 2 which is going to be approximately if we take it to our calculator. So 10 to the parentheses negative 1 divided by 5 close the parentheses and that whole thing needs to be divided by 2 and we find out that x is approximately 0.32. Now again if you ever get to a case where you aren't sure, looking and thinking, I'm not sure that was really right. You can always plug it back in. So we're saying that 5 times the log of 2 times 0.32 plus 4 and that's going to be approximately 3. Remember we rounded this number so it won't be exactly 3. But let's plug and chug and see what we get. 5 times the log of 2 times 0.32 my fault. Close the parentheses and then outside we add 4 and we get 3.03 which is approximately 3. So we know that x really is about 0.32. We're doing this problem. We're ready to just, it's got an involved argument here. What's inside the parentheses is rather involved. But remember it's a log so we have a base of 10. Hop across the pond. We have 10 cubed. Hop back across the pond to see what it's equal to which is 2x minus 7. And then peel the layers again because now we're trying to solve for x. So 10 to the third and this was minus 7 so we're going to add 7 to it is equal to 2x. And then we're going to divide everything by 2. So 10 to the third plus 7 divided by 2 will be equal to x. So 10, care it 3 and then make sure we come out of the exponent plus 7 and I'm going to do this in two parts. So plus 7 gives me this divided by 2. And I get 503.5. Again to check it, if I really needed to check it I could ask for the log of 2 times 503.5 minus 7 equal to 3. So log of 2 times 503.5 and then minus 7 to multiply and then subtract 7 and I get up 3.