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Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | We're almost there. We've almost solved for W. To finish it up, we just have to divide both sides of this equation by 2. And the whole reason why I'm dividing both sides of this equation by 2 is to get rid of this 2 coefficient. This 2 that's multiplying W. So if you divide both sides of this equation by 2, once again,... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | This 2 that's multiplying W. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything. So this is just going to b... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | So we're done. If we flip these two sides, we have our W will be equal to this thing over here. Equals P minus 2L all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this. So let me square this off because this is completely the correct answer. |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this. So let me square this off because this is completely the correct answer. This is the correct answer. But there's other ways that you might have been able to get this answer or other expressions for this answer. You ... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | So let me square this off because this is completely the correct answer. This is the correct answer. But there's other ways that you might have been able to get this answer or other expressions for this answer. You might have also another completely legitimate way to do this problem. Let me write it this way. So our or... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | You might have also another completely legitimate way to do this problem. Let me write it this way. So our original problem is P is equal to 2L plus 2W. Is on this right-hand side. What if we factor out a 2? So let me make this clear. You have a 2 here and you have a 2 here. |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | Is on this right-hand side. What if we factor out a 2? So let me make this clear. You have a 2 here and you have a 2 here. So you can imagine undistributing the 2. So we would get P is equal to 2 times L plus W. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2 so... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | You have a 2 here and you have a 2 here. So you can imagine undistributing the 2. So we would get P is equal to 2 times L plus W. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2 so that we get rid of this 2 on the right-hand side. So if you divide both sides of ... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | Now, we can divide both sides of this equation by 2 so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out. 2 times anything divided by 2 is just going to be the anything. It's equal to P over 2. So let me just rewrite this over here. Le... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | It's equal to P over 2. So let me just rewrite this over here. Let me just rewrite this. So we will get P over 2 is going to be equal to L plus W. And then if we want to solve for W, we just subtract L from both sides. And sometimes you can write it in a separate line like this. Sometimes you can just write it like thi... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | So we will get P over 2 is going to be equal to L plus W. And then if we want to solve for W, we just subtract L from both sides. And sometimes you can write it in a separate line like this. Sometimes you can just write it like this. You can say, I'm going to subtract an L on that side. If I do it on that side, I have ... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | You can say, I'm going to subtract an L on that side. If I do it on that side, I have to do it on this side too. That's the same thing as adding a negative L. And so the right-hand side, you're just left with a W. And then the left-hand side, it could be a negative L plus P over 2, or you could just change the order. A... |
Example Solving for a variable Linear equations Algebra I Khan Academy.mp3 | And you can write this as P over 2 minus L. And this is also an equally legitimate answer. And you're probably saying, hey Sal, wait, these things look different. P minus 2L over 2, that looks different than P over 2 minus L. And they're not. Think about this. We could rewrite this as P over 2 minus 2L over 2. If I hav... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | The two things I'm going to graph are y is equal to 2 to the x power and y is equal to the log base 2 of x. I encourage you to pause the video, make a table for each of them and try to graph them on the same graph paper and see how they are related. If you see how they are related, think about why they are related that... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | I'll make a little table here, different x values and the corresponding y values. x and y, we can start with negative 2, negative 1, 0, 1, 2, 3. In each case, y is going to be 2 raised to this power. 2 to the negative 2 power is going to be 1 fourth. 2 to the negative 1 power is 1 half. 2 to the 0 power is 1. 2 to the ... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the negative 2 power is going to be 1 fourth. 2 to the negative 1 power is 1 half. 2 to the 0 power is 1. 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. Let's graph that. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. Let's graph that. 2 to the third power is 8. 2 to the third power is 8. 2 to the second power is 4. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | Let's graph that. 2 to the third power is 8. 2 to the third power is 8. 2 to the second power is 4. 2 to the first power is 2. 2 to the zeroth power is 1. 2 to the negative 1 power is 1 half. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the second power is 4. 2 to the first power is 2. 2 to the zeroth power is 1. 2 to the negative 1 power is 1 half. 2 to the negative 2 power is 1 fourth. 2 to the negative third power is 1 eighth. It's going to look something like this. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the negative 1 power is 1 half. 2 to the negative 2 power is 1 fourth. 2 to the negative third power is 1 eighth. It's going to look something like this. The graph is going to look something like this right over here. It's kind of your classic, sometimes this will be called your exponential hockey stick. It kind o... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | It's going to look something like this. The graph is going to look something like this right over here. It's kind of your classic, sometimes this will be called your exponential hockey stick. It kind of looks like a hockey stick. It just starts kind of slow and just shoots straight up. Notice, as we go to the left, as ... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | It kind of looks like a hockey stick. It just starts kind of slow and just shoots straight up. Notice, as we go to the left, as x becomes more and more negative, our value approaches 0 but never quite gets there. If we have 2 to the negative 1 millionth power, it's going to be a very, very small number, very, very clos... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | If we have 2 to the negative 1 millionth power, it's going to be a very, very small number, very, very close to 0. But it's not going to be quite 0. We're going to have a horizontal asymptote at y is equal to 0, or the x-axis is a horizontal asymptote. Fair enough. Now let's graph y is equal to log base 2 of x. Before ... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | Fair enough. Now let's graph y is equal to log base 2 of x. Before I graph that, let's just think about another way of representing it. This literally says for any x, what power, what exponent y, if I raise 2 to that, would give me x? This is an equivalent statement as saying 2 to the y power is equal to x. If you noti... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | This literally says for any x, what power, what exponent y, if I raise 2 to that, would give me x? This is an equivalent statement as saying 2 to the y power is equal to x. If you notice what we've done here, between these two things, you're essentially just switching the x's and the y's. Here it's 2 to the x power is ... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | Here it's 2 to the x power is equal to y. Here it's 2 to the y power is equal to x. Really, this and this, you've swapped the x's and the y's. You've swapped the x's and the y's. What we will see is that we can essentially swap these two columns, x and y. Let me just do 1 fourth, 1 half, 1, 1, 2, 4, and 8. Here now we'... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | You've swapped the x's and the y's. What we will see is that we can essentially swap these two columns, x and y. Let me just do 1 fourth, 1 half, 1, 1, 2, 4, and 8. Here now we're saying if x is 1 fourth, what power do we have to raise 2 to to get to 1 fourth? We have to raise it to the negative 2 power. 2 to the negat... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | Here now we're saying if x is 1 fourth, what power do we have to raise 2 to to get to 1 fourth? We have to raise it to the negative 2 power. 2 to the negative 1 power is equal to 1 half. 2 to the 0 power is equal to 1. 2 to the first power is equal to 2. 2 to the second power is equal to 4. 2 to the third power is equa... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the 0 power is equal to 1. 2 to the first power is equal to 2. 2 to the second power is equal to 4. 2 to the third power is equal to 8. Notice all we did is we essentially swapped these two columns. Let's graph this. When x is equal to 1 fourth, y is equal to negative 2. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | 2 to the third power is equal to 8. Notice all we did is we essentially swapped these two columns. Let's graph this. When x is equal to 1 fourth, y is equal to negative 2. When x is 1 half, y is equal to negative 1. When x is 1, y is 0. When x is 2, y is 1. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | When x is equal to 1 fourth, y is equal to negative 2. When x is 1 half, y is equal to negative 1. When x is 1, y is 0. When x is 2, y is 1. When x is 4, y is 2. When x is 8, y is 3. It's going to look like this. |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | When x is 2, y is 1. When x is 4, y is 2. When x is 8, y is 3. It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two? |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two? You'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think a... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | What would you have to reflect about to get these two? You'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think about it, if you swap the axis, you would get the other graph, which is essentially what we're doing. Notice it's symmetric about that line. That's because these are ... |
Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3 | Notice it's symmetric about that line. That's because these are essentially the inverse functions of each other. One way to think about it is we swapped the x's and y's. Just as x becomes more and more negative, you see y approaching 0. Here you see as y is becoming more and more negative, x is approaching 0. Or you co... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So this red curve is the graph of f of x, and this blue curve is the graph of g of x. And I want to try to express g of x in terms of f of x. And so let's see how they're related. So we pick any x, and we could start right here at the vertex of f of x. And we see that at least at that point, g of x is exactly one highe... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So we pick any x, and we could start right here at the vertex of f of x. And we see that at least at that point, g of x is exactly one higher than that. So g of 2, I could write this down, g of 2 is equal to f of 2 plus 1. Let's see if that's true for any x. So then we can just sample over here. Well, let's see. f of 4... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | Let's see if that's true for any x. So then we can just sample over here. Well, let's see. f of 4 is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is one more than that. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | f of 4 is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is one more than that. So it looks like if we pick any point over here, even though there's a little bit of an optical illusion, it looks like they get closer together. They do if you try to find the closest distance between the two. ... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | g of 6 is one more than that. So it looks like if we pick any point over here, even though there's a little bit of an optical illusion, it looks like they get closer together. They do if you try to find the closest distance between the two. But if you look at vertical distance, you see that it stays a constant. It stay... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | But if you look at vertical distance, you see that it stays a constant. It stays a constant 1. So we can actually generalize this. This is true for any x. g of x is equal to f of x plus 1. Let's do a few more examples of this. So right over here, here is f of x in red again. And here is g of x. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | This is true for any x. g of x is equal to f of x plus 1. Let's do a few more examples of this. So right over here, here is f of x in red again. And here is g of x. And so let's say we picked x equals negative 4. This is f of negative 4. And we see g of negative 4 is 2 less than that. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | And here is g of x. And so let's say we picked x equals negative 4. This is f of negative 4. And we see g of negative 4 is 2 less than that. And we see whatever f of x is, g of x, no matter what x we pick, g of x seems to be exactly 2 less. g of x is exactly 2 less. So in this case, very similar to the other one, g of ... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | And we see g of negative 4 is 2 less than that. And we see whatever f of x is, g of x, no matter what x we pick, g of x seems to be exactly 2 less. g of x is exactly 2 less. So in this case, very similar to the other one, g of x is going to be equal to f of x. But instead of adding, we're going to subtract 2 from f of ... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So in this case, very similar to the other one, g of x is going to be equal to f of x. But instead of adding, we're going to subtract 2 from f of x. f of x minus 2. Let's do a few more examples. So here we have f of x in red again. I'll relabel it. f of x. And here is g of x. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So here we have f of x in red again. I'll relabel it. f of x. And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of 3. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of 3. So that, or f of negative 3, I should say. This is negative 3. This is the point negative 3, f of 3. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | Let's say we have in red here, this point right over there is the value of f of 3. So that, or f of negative 3, I should say. This is negative 3. This is the point negative 3, f of 3. So negative 3, f of 3. Now g hits that same value when x is equal to negative 1. When x is equal to negative 1. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | This is the point negative 3, f of 3. So negative 3, f of 3. Now g hits that same value when x is equal to negative 1. When x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. f of negative 3 is equal to f of negative 3. And we could do that with a bunch of points. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | When x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. f of negative 3 is equal to f of negative 3. And we could do that with a bunch of points. We could see that g of 0, which is right there. Let me do it in a color you can see. g of 0 is equivalent to f of negative 2. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | And we could do that with a bunch of points. We could see that g of 0, which is right there. Let me do it in a color you can see. g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of,... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of, well, it's going to be 2 less than x. So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. It's important to realize here, wh... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. It's important to realize here, when I did f of x minus 2 here, and remember, the function is being evaluated. This is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a littl... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | It's important to realize here, when I did f of x minus 2 here, and remember, the function is being evaluated. This is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counterintuitive unless you go through this exercise right over here. So g of... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | When I subtract the 2, this is shifting the function to the right, which is a little bit counterintuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2, we would have actually shifted f to the left. Now let's think about this one. This one seems kind ... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image, but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see. It gets to about 2 there. Then it gets pretty close to 1 right over there. And then it gets about right over there. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | So let's see. It gets to about 2 there. Then it gets pretty close to 1 right over there. And then it gets about right over there. So if I were to take its mirror image, it looks something like this. Its mirror image, if I were to reflect it across the x-axis, it looks something like this. It looks something like this. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | And then it gets about right over there. So if I were to take its mirror image, it looks something like this. Its mirror image, if I were to reflect it across the x-axis, it looks something like this. It looks something like this. So this right over here, we would call. So if this is g of x, when we flip it that way, t... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | It looks something like this. So this right over here, we would call. So if this is g of x, when we flip it that way, this is the negative g of x. When x equals 4, g of x looks like it's, I don't know, about negative 3 and 1 half. You take the negative of that, you get positive. I guess it should be closer. You get pos... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | When x equals 4, g of x looks like it's, I don't know, about negative 3 and 1 half. You take the negative of that, you get positive. I guess it should be closer. You get positive 3 and 1 half if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us there. It looks like we act... |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | You get positive 3 and 1 half if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us there. It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3. |
Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3 | It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3. So it looks like this red graph right over here is 3 times this graph. So this is 3 times negative g of x, which is equal to negative 3 g of x. So her... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Let's learn a little bit about the wonderful world of logarithms. So, we already know how to take exponents. If I were to say 2 to the 4th power, what does that mean? Well, that means 2 times 2 times 2 times 2. 2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, tim... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, that means 2 times 2 times 2 times 2. 2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, times 2 is 16. But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know w... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is. So, for example, let's say that I start with 2, and I say I'm raising it to some power. What does that power have to be to get 16? Well, we just f... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | What does that power have to be to get 16? Well, we just figure that out. X would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let m... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful. Log base 2, so I'll do this 2 in blue, log base 2 of 16 is... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Or is equal in this case, since we have the x there, is equal to x. This and this are completely equivalent statements. This is saying, hey, well, if I take 2 to some x power, I get 16. This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you've got to raise it to the fourth power. Once again, x is equal to 4. So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log base 3 of 81. What would this evaluate to? Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting ... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81. You had to use algebra here. Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra. We didn't have to set it equal to x. We could just say this evaluates to the power I need to raise 3 to to get to 81. Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit. So 3 to the... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit. So 3 to the first power is just 3. 3 to the second power is 9. 3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4. |
Logarithms Logarithms Algebra II Khan Academy.mp3 | 3 to the second power is 9. 3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hop... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | 3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number. Let's say we want to take log base 6 of 216... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number. Let's say we want to take log base 6 of 216. What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to t... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to the second power is 36. 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. |
Logarithms Logarithms Algebra II Khan Academy.mp3 | 6 to the second power is 36. 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216. |
Logarithms Logarithms Algebra II Khan Academy.mp3 | So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216. Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to? |
Logarithms Logarithms Algebra II Khan Academy.mp3 | 6 to the third power is equal to 216. Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to r... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the si... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | 8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending ... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending on how you view it. What is log base 100 of 1? Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about ... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | What is log base 100 of 1? Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about it. I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 1... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | That's one way to think about it. I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation. So if I set this to be equal to x, this is literally saying 100 to ... |
Logarithms Logarithms Algebra II Khan Academy.mp3 | Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation. So if I set this to be equal to x, this is literally saying 100 to what power is equal to 1? Well, anything to the zeroth power is equal to 1. So in this case, x is equa... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | All right, complete the following sentence about the daily percent change in the mass of the sunfish. Every day there is a blank percent addition or removal from the mass of the sunfish. So one thing that we know almost from the get-go, we know that the sunfish gains weight. And we also see that as t grows, as t grows,... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | And we also see that as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one, m of t is going to grow. So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day. Well, let's think a... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | But let's think about how much is added every day. Well, let's think about it. Well, let's see if we can rewrite this. This is, I'm gonna just focus on the right-hand side of this expression. So 1.35 to the t over six plus five, that's the same thing as 1.35 to the fifth power times 1.35 to the t over sixth power. And ... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | This is, I'm gonna just focus on the right-hand side of this expression. So 1.35 to the t over six plus five, that's the same thing as 1.35 to the fifth power times 1.35 to the t over sixth power. And that's going to be equal to 1.35 to the fifth power times 1.35, and I can separate this t over six as 1 sixth times t. ... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass and multiply it by this common ratio. The common ratio here isn't, the way I've written it, isn't 1.35. It's 1.35 to the 1 sixth power. Let me draw a little table here to make that really, really clea... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | It's 1.35 to the 1 sixth power. Let me draw a little table here to make that really, really clear. And all of that algebraic manipulation I just did is just so I could simplify this so I have some common ratio to the t power. So t and m of t. So based on how I've just written it, when t is zero, well, if t is zero, thi... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | So t and m of t. So based on how I've just written it, when t is zero, well, if t is zero, this is one, so then we just have our initial mass is going to be 1.35 to the fifth power. And then when t is equal to one, when t is equal to one, it's gonna be our initial mass, 1.35 to the fifth power times our common ratio, t... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | And so every day, well, let me get, every day we are growing, every day we are growing by our common ratio, 1.35 to the 1 sixth power. Actually, let me get a calculator out. We're allowed to use calculators in this exercise. So 1.35 to the, to the, open parentheses, one divided by six, close parentheses, power is equal... |
Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3 | So 1.35 to the, to the, open parentheses, one divided by six, close parentheses, power is equal to 1.05, I'll say 1.051 approximately. So this is approximately 1.051. So we could say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing by a factor of 1.051. Well, growing by a factor of... |
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