Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. When using a calculator to find the following common and natural logarithms, we need to 1st understand that if it says log of two, there's an understood #10 for base 10. That base we use all the time. So log of 2 is really the same thing as log base 1:50 and if we put that into our calculator we get .3010 approximately. We want a round of 4 decimal places. Log 8200 once again means log base 10 of 8200, and when we put that into our calculator we get approximately 3.9138. The natural log of 12 lane stands for natural log and we can think of that as log base E where E is the natural number. It's a number that occurs frequently in nature, hence the name the natural number. So log base E of 12. On your calculators, it frequently will say lane is approximately 2.4849, east is the natural number or the base to the 3.06. And when I put that into my calculator, I get 21 point 327. And this time we're going to round up to the six because it was 55. Whenever there's a 5 or higher, we round the digit to immediately to the left up. So on this next type of problem, it says find each of the logarithms using the change of base formula. And depending on what calculator you use, some calculators nowadays will actually allow you to type this in just like it is with a log and a base six and a 37. If your calculator will allow you to do that, you would get 2.0153 because it's 29 and that nine means to round the two up to three. If you're using a calculator that doesn't have the ability to do a different base, we use a formula called the change a base formula and we change into either base 10 or base E. So log base 6 of 37 could really be thought of as log base 10 of 37 over log base 5:50. Now that really is the same thing as the natural log of 37 over the natural log of 6. If you don't believe me, type those in your calculators and show to yourself that those are the same. Now I want to actually show why this is true. If I come over here and we say log base 6 of 37, let's let it equal some X. Then if I change that into exponential form, I get 6 to the X equal 37. Now we could take the log of each side and it doesn't matter if we take the natural log or if we take the log base 10. If we take the natural log of each side that X because it's in the exponent by the power rule I'm going to bring down in front. So XLN 6 equal LN37. And if my goal is to get X by itself, we divide each side by L and six. So X equal L and 37 / L and six would have worked exactly the same way if I had taken log base 10. So once again, if your calculator will take it with a base, so log base 25 of 625, that's just going to give us the number 2. If your calculator can't do that, change it to base 10, log base 10 of 625 over log base 10 of 25, or the natural log of 625 over the natural log of 25. When you type those in, you'll get to either direction for the next type of problem. Here we want to graph and state the domain and range of each. So when we graph this, we're going to realize that it's going to look really similar to a graph that we've already seen with the exponentials in the bases. If you recall, when we graphed Y equal 2 to the X, we had a graph that looked something really rough. Sketch like this because 2:00 to the second power would have been four, and two to the first would have been two, and two to the zero would have been one, and two to the -1 would have been 1/2 and two to the -2 a fourth. So we would have gotten closer and closer and closer to the asymptote that we put in as a dotted line. The same thing is going to happen with E to the X and we're going to use our calculator to type this in and we're just going to put in a rough sketch. So our rough sketch we're going to get close to the asymptote, which is going to occur at the X axis or with the line y = 0 and it's going to go up like. So we know that this point is going to be the point when X IS0Y is 1 because anything to the 0 power is 1, so E to the zero is 1. The exception to that is 0 to the zero which is undefined. So the domain here we can see is going to be the all real numbers or negative Infinity to Infinity and the range is only going to be positive Y values so zero to Infinity. Now what happens when we put a + 2 at the end? That +2 really just takes that whole graph and it shifts the whole graph up to so it also shifts. When I say the whole graph I mean everything. It also shifts the asymptote up by two. So once again a very rough graph. So this point now is when X is 0 we get Y being 3 because E to the zero was one and 1 + 2 is 3. So our domain again is going to be all real numbers, but our range this time is going to be from 2 and above, or two to Infinity because our asymptote moved up by two. If we look at the next grouping of problems, E to the X -, 3, what's the -3 do? It's going to take the original graph and it's going to shift everything down by three. So my asymptote by the way, which is a line, it's the line Y equal -3. Should go back and put that in this one is Y equal to y = 0. One way that I personally think about it is it's whatever the number at the end plus or minus is when it's E to a power. If there isn't a number, it's understood to be a + 0, so that's a +2, so Y equal 2 -, 3, so Y equal -3. When we sketch in the graph, it's going to look something like this. When X was 0, we have our Y being -2 because E to the zero was one. 1 -, 3 is -2. The domain once again is going to be all reals. The range this time is going to be from -3 to Infinity. We get closer and closer and closer to -3 and then we go up to positive Infinity. What's the two in front do? It's going to make the graph stretch in the Y direction by a factor of 2. And what's the negative .5 going to do? It's going to make our X's compress. So when we graph this, and once again, we're supposed to be using our calculators to help us see all of this. So if we have a 2 * E to the negative .5, the negative is actually going to reflect it also. So we're going to get a graph that's going to look like this. Now there was no plus or minus number after the E terms, so our asymptote is going to be y = 0 because there was no number that was plus or minus. If we thought about putting X as 0 zero times anything is 0. E to the zero is 1 and 2 * 1, so 02. When we look at our graph, we realize the domain is all real numbers and the range to this time is 0 to Infinity. Now switching to do a natural log graphs. The natural log is really, really just the inverse of the natural number. So we're going to expect that these graphs are going to look like a logarithmic graph. So when we look at this graph, we have lane X. So we're going to have a regular lane graph, but then we're going to +3, and what's that +3 going to do? Plus 3 is going to take that whole graph and it's going to shift it up three. So a regular lane X graph before any moving or changing would have an asymptote at X = 0. And it would look a lot like this, so that when X is one, our Y is 0 because the natural log of one is 0. So this plus three, we're going to still have our asymptote. It's not going to change because we're moving it up and down. But the asymptote this time is a horizontal line. Nope, vertical line. Sorry, I always get those mixed up in my own head. So here. Now when X is 0 or X is one, log of one is zero, 0 + 3 is 3. So we get the .13. Looking at this graph, we can see the domain is going to be 0 to Infinity, and the range is going to be all reals, because this is going to keep getting a little bigger and a little bigger and a little bigger as we go out to Infinity. The natural log of X + 2, this +2, because it's in parentheses, is actually going to shift our graph right and left this time. And if we think about it, it's going to shift X + 2, so 2 to the left. The easiest way I can tell you to do that is to think about setting that X + 2 equal to 0 and solving for X to see the shift. So our new asymptote line is going to be X equal -2. If I stick in -1 for X -1 + 2 is 1 and the natural log of one is 0. So we know that the point -1 zero when we look at this, the domain's going to be all reals again, our range this time. Nope, I said that backwards. Sorry, sorry sorry. Our domain, our X values are going to be -2 to Infinity because our vertical asymptote shifted to to the left. Our range is negative Infinity to Infinity. I hope these examples helped. Thank you and have a wonderful day.