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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 fourth 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, times 2 is 16. But what if we think about things in another way. We know that we get to 16 when we raise 2 to some power but 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 figured 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 me make it a little bit more colourful. Log, base 2-- I'll do this 2 in blue... Log, base 2, of 16 is equal to what, 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'll say, "well you have to raise it to the fourth power and once again 'x' is equal to 4. So with that out of the way let's try more examples of evaluating logarithmic expressions. Let's say you had... log, base 3, of 81. What would this evaluate to? Well this is a reminder, this evaluates to the power we have to raise 3 to, to get to 81. So if you want to, you can set this 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 properties later on. But you didn't necessarily have to use algebra. To do it this way, to say that 'x' is the power you raise 3 to to get to 81, you had to use algebra here, while with just a straight up logarithmic expression, you didn't really have to use any algebra, we didn't have to say that it was equal to 'x', we could just say that this evaluates to the power I need to raise 3 to to get to 81. 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 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 four. So we could say... Log, base 3, of 81, is equal to-- I'll do this in a different colour. 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 hopefully you'll get the hang of it. So let's try a 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 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. 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, I dunno, log, base 2, of 64. So what does this evaluate to? Well once again we're asking ourselves, "well this will evaluate to the exponent that I have to raise this base 2, 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. 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 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? I'll let you think about that for a second. 100 is a subscript so it's, log, base 100, of 1. That's one way to think about it, I'll put 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 100 to, to get to 1?" So let me write this down as an equation. If I set this to be equal to 'x', this is literally saying 100, to what power, is equal to 1? Well anything that a 0 power is equal to 1. So in this case 'x' is equal to 0. So log, base 100, of 1, is going to be equal to 0. Log base anything of 1, is going to be equal to 0 because anything to the 0 power and we're not talking about 0 here. Anything that is to the power of 0 that is not 0, is going to be equal to 1.

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Intro to logarithms (video) | Logarithms | Khan Academy

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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 fourth power, what does that mean? Well that means 2 times 2 times 2 times 2.

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- [Voiceover] We're told the graph of y equals two to the x is shown below, alright. Which of the following is the graph of y equals two to the negative x minus five? So there's two changes here. Instead of two to the x, we have two to the negative x and then, we're not leaving that alone, we, then, subtract five. So let's take 'em step by step. So let's first think about what y equals two to the negative x would look like. Well, any input we now put into an x, we're now going to take the negative of. So if I input a two, it's like taking the opposite of the two and then, inputting that into two to the x. And so, what we're essentially going to do is flip this graph over the y-axis. So, here we have the point two comma four. Over here, we're going to have the point negative two comma four. When x is zero, they're going to give us the same value. So they're both going to have the same y-intercept. And so are graph is going to look like, our graph is going to look something like, this. They're going to be mirror images flipped around the y-axis. So, it's going to look like that. That is the graph of y is equal to two to the negative x. And then we have to worry about the subtracting five from it. Well, that's, you're subtracting five from your final y-value so that's going to work. You're subtracting five to get your y-value now, where your y-value is going to be five lower, is I guess the best way to say it, so this is going to shift the graph down by five. So instead of having the y-intercept there, it's going to be five lower. One, two, one, each hash mark is two so this is one, two, three, four, five, is going to be right over there. So shift down by five, two, four, five. It's going to look like that. And the asymptote, instead of the asymptote, going towards y equals zero, the asymptote is going to be at y is equal to negative five. So the asymptote is going to be y equals negative five. So it should look something like, something like what I am drawing right now. Something like, something like that. So now we can look at which choices. So this should be the graph of y equals two to the negative x minus five. So let's see which of these choices depict that. So this first choice actually seems to be spot-on. It's exactly what we drew. Well, we can look at the other ones, just in case. Well, this is, this looks like, what did they do over here? Looks like they, instead of flipping over the y-axis, they took the, they flipped over the x-axis and then they shifted down, so that's not right. Here it looks like they got what we got but they flipped it over the x-axis. And this looks like they flipped it over the y-axis but then they shifted, instead of shifting down by five, it looks like they shifted to the left by five. So we should feel pretty good especially 'cause we essentially drew this before even looking at the choice.

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Transforming exponential graphs (video) | Khan Academy

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Voiceover] We're told the graph of y equals two to the x is shown below, alright. Which of the following is the graph of y equals two to the negative x minus five? So there's two changes here. Instead of two to the x, we have two to the negative x and then, we're not leaving that alone, we, then, subtract five.

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𝑒 and compound interest (video) | Khan Academy

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Sal introduces a very special number in the world of math (and beyond!), the constant 𝑒.

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Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties. But what I'm going to do is I'm going to show you the properties. And then show you how they can be used. It's going to be little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a-- Oh that's not the right. Let's see. I want to change-- There you go. Let's say I say that a-- Let me start over. a to the b is equal to c. So if we-- a to the b power is equal to c. So another way to write this exact same relationship instead of writing the exponent, is to write it as a logarithm. So we can say that the logarithm base a of c is equal to b. So these are essentially saying the same thing. They just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power you get c. And then you figure out what b is. So they're the exact same relationship, just stated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm-- Let me do a more cheerful color. The logarithm, let's say, of any base-- So let's just call the base-- Let's say b for base. Logarithm base b of a plus logarithm base b of c-- and this only works if we have the same bases. So that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, well I don't know, examples. So this is saying that-- I'll switch to another color. Let's make mauve my-- Mauve-- I don't know. I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of-- I don't know --of 8 plus logarithm base 2 of-- I don't know let's say --32. So, in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Well let's see if that's true. Just trying out this number and this is really isn't a proof. But it'll give you a little bit of an intuition, I think, for what's going on around you. So log-- So this is-- We just used our property. This little property that I presented to you. And let's just see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see. 2 to the fourth power is 16. 2 to the fifth power is 32. So this right here is 2 to the-- This is 5, right? And 2 to the what power is equal to 256? Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent prop-- I don't know. I don't know the names of things. And that equals 2 to 8, 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes the logarithms interesting is and why-- It's a little confusing at first. And you can watch the proofs if you really want kind of a rigorous-- my proofs aren't rigorous. But if you want kind of a better explanation of how this works. But this should hopefully give you an tuition for why this property holds, right? Because when you multiply two numbers of the same base, right? Two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. So let's do a-- Let me show you another log property. It's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of-- well I ran out. I'm running out of space --a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of-- I don't know --log base 3 of-- well you know, let's make it interesting --log base 3 of 1/9 minus log base 3 of 81. So this property tells us-- This is the same thing as-- Well I'm ending up with a big number. Log base 3 of 1/9 divided by 81. So that's the same thing as 1/9 times 1/81. I used two large numbers for my example, but we'll move forward. So let's see. 9 times 8 is 720, right? 9 times-- Right. 9 times 8 is 720. So this is 1/729. So this is log base 3 over 1/729. So what-- What does-- 3 to what power is equal to 1/9? Well 3 squared is equal to 9, right? So 3-- So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1/9, right? The negative just inverts it. So this is equal to negative 2, right? And then minus-- 3 to what power is equal 81? 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to-- Well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus sixth power is equal to 1/729. So that's my question. Is 3 to the minus sixth power, is that equal to 7-- 1/729? Well that's the same thing as saying 3 to sixth power is equal to 729, because that's all the negative exponent does is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here. But let's see. 3 to the third power-- This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video. In the next video, I'll introduce you to the last two logarithm properties. And, if we have time, maybe I'll do examples with the leftover time. I'll see you soon.

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Intro to logarithm properties (1 of 2) (video) | Khan Academy

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Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties.

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Voiceover:What I want to do in this video is graph up a classic exponential function and then graph a related logarithmic function and see how the two are related visually. The two things I'm going to graph are y is equal to two to the x power and y is equal to the log base two 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. See how they are related and if you see how they're related, think about why they are related that way. Let's first start with y equals two to the x power. I'm going to make a little table here, different x values and the corresponding y values. x and y, we can start with negative two, negative one, zero, one, two, three. In each case y is going to be two raised to these power. Two to the negative two power is going to be 1/4. Two to the negative one power is 1/2. Two to the zero power is one. Two to the first power is two. Two to the second power is four. Two to the third power is eight. Let's graph that. Two to the third power is eight. Two to the second power is four. Two to the first power is two. Two to the zeroth power is one. Two to the negative one power is 1/2. Two to the negative two power is 1/4. Even the two to the negative third power is going to be 1/8, so 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 because it kind of looks like a hockey stick where it just kind of starts kind of slow and just oohh bam, shoots straight up. Notice as we go to the left as x becomes more and more and more negative our value approaches zero but never quite gets there. If we have two to the negative one millionth power it's going to be a very, very small number, very, very close to zero but it's not going to be quite zero. We're going to have a horizontal asymptote at y is equal to zero or the x-axis is a horizontal asymptote. Fair enough. Now let's graph y is equal to log base two 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 two to that would give me x. This is an equivalent statement as saying two 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's two to the x power is equal to y. Here's two to the y power is equal to x. Really this and this 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, so let me just do 1/4, 1/2, one, two, four, and eight. Here now we're saying if x is 1/4, what power do we have to raise two to, to get to 1/4. We have to raise it to the negative two power. Two to the negative one power is equal to 1/2. Two to the zero power is equal to one. Two to the first power is equal to two. Two to the second power is equal to four. Two to the third power is equal to eight. Notice all we did, as we essentially swapped these two columns, so let's graph this. When x is equal to 1/4, y is equal to negative two. When x is 1/2, y is equal to negative one. When x is one, y is zero. When x is two, y is one. When x is four, y is two. When x is eight, y is three. 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? Well 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, if you swap the axis you would get the other graph. It's essentially what we're doing. Notice it's symmetric about that line and 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 this, as x becomes more and more and more and more negative you see y approaching zero. Here you see is y is becoming more and more negative as x is approaching zero, or you could say as x approaches zero y becomes more and more and more negative. The whole point of this is just to give you an appreciation for the relationship between an exponential function and a logarithmic function. They're essentially inverses of each other. You see that in the graphs, they're reflections of each other about the line y is equal to x.

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Graphical relationship between 2ˣ and log₂(x) (video) | Khan Academy

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Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

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- [Voiceover] Carlos has taken an initial dose of a prescription medication. The relationship between the time, between the elapsed time t, in hours, since he took the first dose and the amount of medication, M of t, in milligrams, in his bloodstream is modeled by the following function. Alright, in how many hours will Carlos have 1 milligram of medication remaining in his bloodstream? So M of what t is equal to, so we need essentially to solve for M of t is equal to 1 milligram. Because M of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So M of t is, they give us a definition, it's model is an exponential function, 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20 and so we will get e to the negative 0.8 t is equal to one over 20, one over 20. Which we could write as 0.05, 0.05. I have a feeling we're gonna have to deal with decimals here regardless. And so how do we, how do we solve this? Well one way to think about it, one way to think about it if we took, what happens if we took the natural log of both sides? And just a remember, a reminder, the natural log is the logarithm base e. So actually let me write this, let me write this a little bit differently. So this is zero and that is 0.05. So I'm gonna take the natural log of both sides, so ln, ln. So the natural log, this says, what power do I have to raise e to, to get to e to the negative 0.8 t? Well I've got to raise e to the, this simplifies to negative 0.8 t. Once again natural log this thing... let me clarify ln of e to the negative 0.8 t. This is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to, to get to e to the negative 0.8 t? Well I have to raise it to the negative 0.8 t power. So that's why the left-hand side simplified to this and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, the natural log of 0.05 all of that and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8 and so t is going to be equal to all of this business. On the left-hand side now we just have a t and on the right-hand side we have all of this business which I think a calculator will be valuable for. So let me get a calculator out. Clear it out and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log. We get that value and we want to divide it by negative .8. So divided by, divided by .8 negative. So we're gonna divide by .8 negative is equal to, let's see they wanted us to round to the nearest hundredth so 3.74, so it'll take 3.74, seven four hours for his dosage to go down to one milligram where it actually started at 20 milligrams. When t equals zero it's 20 after 3.74 hours he's down in his bloodstream to one milligram. I guess his body has metabolized the rest of it in some way.

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Exponential model word problem: medication dissolve (video) | Khan Academy

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Voiceover] Carlos has taken an initial dose of a prescription medication. The relationship between the time, between the elapsed time t, in hours, since he took the first dose and the amount of medication, M of t, in milligrams, in his bloodstream is modeled by the following function.

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Creating Tables of Exponential Functions

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Creating Tables of Exponential Functions

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- [Voiceover] g is an exponential function with an initial value of -2. So, an initial value of -2, and a common ratio of 1/7, common ratio of 1/7. Write the formula for g(t). Well, the fact that it's an exponential function, we know that its formula is going to be of the form g(t) is equal to our initial value which we could call A, times our common ratio which we could call r, to the t power. It's going to have that form. And they tell us what the initial value is. It's -2. So this right over here is -2. And we know that the common ratio is 1/7. So this is 1/7. So let me just write it again a little bit neater. g(t) is going to be equal to our initial value, -2, times times our common ratio, 1/7, to the t power. And hopefully this makes sense. Initial value is this number. Well, if t is equal to 0, then 1/7 to the zero power is 1. And so g(0), you could do that at time as being equal to zero if your t is time, would be equal to -2. So that would be our initial value. And then if you think about it, every time you increase t by one, you're going to mulitply by 1/7 again. And so, the ratio between successive terms is going to be 1/7. And so that's why we call that the common ratio. Hopefully, you found that interesting.

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Writing exponential functions | Algebra (video) | Khan Academy

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Voiceover] g is an exponential function with an initial value of -2. So, an initial value of -2, and a common ratio of 1/7, common ratio of 1/7. Write the formula for g(t).

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Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. And the reason why this is useful is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log-- I'll make x 10-- log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point-- and they want it to the nearest thousandth-- so 2.861. So this is approximately equal to 2.861. And we can verify it because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power. And this number is closer to 3 than it is to 2. Well, let's verify it. So if I take 5 to that power, and then let me type in-- let me just type in what we did to the nearest thousandth-- 5 to the 2.861. So I'm not putting in all of the digits. What do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good. That this is the power that I have to raise 5 to to get to 100. Now with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base a-- I'll try to be fair to the colors-- log base a of b. Let's say I set that to be equal to some number. Let's call that equal to c, or I could call it e for-- Well, I'll say that's equal to c. So that means that a to the c-th power is equal to b. This is an exponential way of writing this truth. This is a logarithmic way of writing this truth. This is equal to b. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this? 10 to the same power will be equal to this, because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. And I'll actually do log base x to prove the general case, here. So I'm going to take log of base x of both sides of this. So this is log base x of a to the c power-- I try to be faithful to the colors-- is equal to log base x of b. And let me close it off with orange, as well. And we know from our logarithm properties, log of a to the c is the same thing as c times the logarithm of whatever base we are of a. And of course, this is going to be equal to log base x of b. Let me put-- I can just write a b, right over there. And if we wanted to solve for c, you just divide both sides by log base x of a. So you would get c is equal to-- and I'll stick to the color-- so it's log base x of b, which is this, over log base x of a. And this was what c was. c was log base a of b. It's equal to log base a of b. Let me write it this way. Let me write it-- Well, let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going, but I want to be fair to the colors. So c is equal to log base x of b over-- let me scroll down a little bit-- log base x, dividing both sides by that, of a. And we know from here I can just copy and paste it, this is also equal to c. This is how we defined it. So let me copy it and then let me paste it. So this is also equal to c. And we're done. We've proven the change of base formula. Log base a of b is equal to log base x of b divided by log base x of a. In this example, a was 5, b is 100, and the base that we switched it to is 10. x is 10.

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Evaluating logarithms: change of base rule (video) | Khan Academy

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Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm.

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- [Voiceover] Let's get some practice solving some exponential equations, and we have one right over here. We have 26 to the 9x plus five power equals one. So, pause the video and see if you can tell me what x is going to be. Well, the key here is to realize that 26 to the zeroth power, to the zeroth power is equal to one. Anything to the zeroth power is going to be equal to one. Zero to the zeroth power we can discuss at some other time, but anything other than zero to the zeroth power is going to be one. So, we just have to say, well, 9x plus five needs to be equal to zero. 9x plus five needs to be equal to zero. And this is pretty straightforward to solve. Subtract five from both sides. And we get 9x is equal to negative five. Divide both sides by nine, and we are left with x is equal to negative five. Let's do another one of these, and let's make it a little bit more, a little bit more interesting. Let's say we have the exponential equation two to the 3x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video, and see if you can tell me what x is going to be, or what x needs to be to satisfy this exponential equation. All right, so you might, at first, say, oh, wait a minute, maybe 3x plus five needs to be equal to x minus seven, but that wouldn't work, because these are two different bases. You have two to the 3x plus five power, and then you have 64 to the x minus seven. So, the key here is to express both of these with the same base, and lucky for us, 64 is a power of two, two to the, let's see, two to the third is eight, so it's going to be two to the third times two to the third. Eight times eight is 64, so it's two to the sixth is equal to 64, and you can verify that. Take six twos and multiply them together, you're going to get 64. This was just a little bit easier for me. Eight times eight, and this is the same thing as two to the sixth power, is 64, and I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64. Let me rewrite the whole thing. So, this is two to the 3x plus five power is equal to, instead of writing 64, I'm going to write two to the sixth power, two to the sixth power, and then that to the x minus seventh power, x minus seven power. And to simplify this a little bit, we just have to remind ourselves that, if I raise something to one power, and then I raise that to another power, this is the same thing as raising my base to the product of these powers, a to the bc power. So, this equation I can rewrite as two to the 3x plus five is equal to two to the, and I just multiply six times x minus 7, so it's going to be 6x, 6x minus six times seven is 42. I'll just write the whole thing in yellow. So, 6x minus 42, I just multiplied the six times the entire expression x minus 7. And so now it's interesting. I have two to the 3x plus 5 power has to be equal to two to the 6x minus 42 power. So these need to be the same exponent. So, 3x plus five needs to be equal to 6x minus 42. So there we go. It sets up a nice little linear equation for us. 3x plus five is equal to 6x minus 42. Let's see, we could get all of our, since, I'll put all my Xs on the right hand side, since I have more Xs on the right already, so let me subtract 3x from both sides. And let me, I want to get rid of this 42 here, so let's add 42 to both sides, and we are going to be left with five plus 42 is 47, is equal to, 47 is equal to 3x. Now we just divide both sides by three, and we are left with x is equal to 47 over three. X is equal to 47 over three. And we are done.

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Solving exponential equations using exponent properties (video) | Khan Academy

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Voiceover] Let's get some practice solving some exponential equations, and we have one right over here. We have 26 to the 9x plus five power equals one. So, pause the video and see if you can tell me what x is going to be.

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Logarithmic equations: variable in the argument

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- [Voiceover] Derek sent a chain letter to his friends, asking them to forward the letter to more friends. The group of people who receive the email gains 9/10 of its size every three weeks, and can be modeled by a function, P, which depends on the amount of time, t in weeks. Derek initially sent the chain letter to 40 friends. Write a function that models the group of people who receive the email t weeks since Derek initially sent the chain letter. So pause the video if you wanna have a go at this. All right, now the way I'd like to think about this, let's just gonna table with values for t and our function P which is a function of t. For some values that we can just pull out of the description here. So when t is zero, when it's been zero weeks since Derek initially sent the chain letter, how many people have gotten it? Well, they tell us. He initially, Derek initially sent the chain letter to 40 friends. So t equals zero. P of t or P of zero is 40. Now, what's an interesting time period? It says that the email, the number of people who receive the email gains 9/10 or increases by 9/10 every three weeks, every three weeks. So after three weeks, so three weeks have gone by. So I'm just adding three to t. What is P of t going to be? Well, they tell us it's going to gain 9/10 of its size. So it's going to be 40 plus 9/10 times 40 which is going to be equal to, what? Well, that's equal to 40. We factor out our 40. 40 times one plus 9/10 or you could say this is equal to 40 times, whoops, 40 times 1.9. Another way of thinking about it, after three weeks, we've grown 90%. That's another way of saying that the number of people who receive the email gains 9/10 of its size. You could say the group of people who receive the email grows 90% every three weeks. And so if we go another three weeks, so plus another three weeks, I could say, well, let me just write this, this is six weeks. Well, how many people would have received the email? It was going to be this number. It's going to be grown another 90%. So we're gonna multiply it times 1.9 again. So it's going to be 40 times 1.9 times 1.9. We're gonna grow by another 9/10. Growing by 9/10 is the same thing as multiplying by one and 9/10. The one is what you already are. Then you're growing by another 9/10. So this is the same thing as 40 times 1.9 squared. We go another three weeks, nine weeks. We're gonna grow another 90%. So you're gonna multiply it by, you're gonna take this number and multiply by 1.9 again which is going to be 1.9 to the third power. And so what's going on over here? Well, we can see some exponential function. We have our initial value. And every three weeks, we're multiplying by 1.9. So 1.9 would be our common ratio. So we could say that P of t is equal to our initial value 40 times our common ratio 1.9. And we multiply by 1.9 every three weeks. So we could just say, how many three week periods have passed by? Well, we will take t and divide it by three. T divided by three is a number of three week periods that have gone by. And there you have it. And notice, t equals zero, 1.9 to the zeroth power is one so 40 times one. T equals three, that's gonna be 1.9 to the first power, three over three. And so we're gonna grow by 90% and so on and so forth. So feeling pretty good about this.

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Constructing exponential models (video) | Khan Academy

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Voiceover] Derek sent a chain letter to his friends, asking them to forward the letter to more friends. The group of people who receive the email gains 9/10 of its size every three weeks, and can be modeled by a function, P, which depends on the amount of time, t in weeks.

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Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

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Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets. For K-12 kids, teachers and parents.

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Pre-Cal Section: All of Section F "Exponential and  Logarithmic Functions"

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Pre-Cal Section: All of Section F "Exponential and  Logarithmic Functions"

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Set students up for success in Precalculus and beyond! Explore the entire Precalculus curriculum: polynomials, derivatives, and more. Try it free!

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Exponential Growth & Decay application problem

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Exponential Growth & Decay application problem

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- [Narrator] What we're going to do in this video is quickly review exponential growth and then use that as our platform to introduce ourselves to exponential decay. So let's review exponential growth. Let's say we have something that, and I'll do this on a table here. Just gonna make that straight. So let's say this is our x and this is our y. Now let's say when x is zero, y is equal to three. And every time we increase x by 1, we double y. So y is gonna go from three to six. If x increases by one again, so we go to two, we're gonna double y again. And so six times two is 12. This right over here is exponential growth. And you could even go for negative x's. When x is negative one, well, if we're going back one in x, we would divide by two. So this is going to be 3/2. 3/2. And notice if you go from negative one to zero, you once again, you keep multiplying by two and this will keep on happening. And you can describe this with an equation. You could say that y is equal to, and sometimes people might call this your y intercept or your initial value, is equal to three, essentially what happens when x equals zero, is equal to three times our common ratio, and our common ratio is, well, what are we multiplying by every time we increase x by one? So three times our common ratio two, to the to the x, to the x power. And you can verify that. Pick any of these. When x is equal to two, it's gonna be three times two squared, which is three times four, which is indeed equal to 12. And we can see that on a graph. So let me draw a quick graph right over here. So, I'm having trouble drawing a straight line. All right, there we go. And let's see. We could go, and they're gonna be on a slightly different scale, my x and y axes. So this is x axis, y axis. And we go from negative one to one to two. Let's see, we're going all the way up to 12. So let's see, this is three, six, nine, and let's say this is 12. We could just plot these points here. When x is negative one, y is 3/2. So looks like that, then at y equals zero, x is, when x is zero, y is three. When x equals one, y has doubled. It's now at six. When x is equal to two, y is 12. And you will see this tell-tale curve. And so there's a couple of key features that we've Well, we've already talked about several of them, but if you go to increasingly negative x values, you will asymptote towards the x axis. It'll never quite get to zero as you get to more and more negative values, but it'll definitely approach it. And as you get to more and more positive values, it just kind of skyrockets up. We always, we've talked about in previous videos how this will pass up any linear function or any linear graph eventually. Now, let's compare that to exponential decay. Exponential, exponential decay. An easy way to think about it, instead of growing every time you're increasing x, you're going to shrink by a certain amount. You are going to decay. So let's set up another table here with x and y values. That was really a very, this is supposed to, when I press shift, it should create a straight line but my computer, I've been eating next to my computer. Maybe there's crumbs in the keyboard or something. (laughs) All right. So here we go. We have x and we have y. And so let's start with, let's say we start in the same place. So when x is zero, y is 3. But instead of doubling every time we increase x by one, let's go by half every time we increase x by one. So when x is equal to one, we're gonna multiply by 1/2, and so we're gonna get to 3/2. Then when x is equal to two, we'll multiply by 1/2 again and so we're going to get to 3/4 and so on and so forth. And if we were to go to negative values, when x is equal to negative one, well, to go, if we're going backwards in x by one, we would divide by 1/2, and so we would get to six. Or going from negative one to zero, as we increase x by one, once again, we're multiplying we're multiplying by 1/2. And so how would we write this as an equation? I encourage you to pause the video and see if you can write it in a similar way. Well, it's gonna look something like this. It's gonna be y is equal to You have your, you could have your y intercept here, the value of y when x is equal to zero, so it's three times, what's our common ratio now? Well, every time we increase x by one, we're multiplying by 1/2 so 1/2 and we're gonna raise that to the x power. And so notice, these are both exponentials. We have some, you could say y intercept or initial value, it is being multiplied by some common ratio to the power x. Some common ratio to the power x. But notice when you're growing our common ratio and it actually turns out to be a general idea, when you're growing, your common ratio, the absolute value of your common ratio is going to be greater than one. Let me write it down. So the absolute value of two in this case is greater than one. But when you're shrinking, the absolute value of it is less than one. And that makes sense, because if the, if you have something where the absolute value is less than one, like 1/2 or 3/4 or 0.9, every time you multiply it, you're gonna get a lower and lower and lower value. And you could actually see that in a graph. Let's graph the same information right over here. And let me do it in a different color. I'll do it in a blue color. So when x is equal to negative one, y is equal to six. When x is equal to zero, y is equal to three. When x is equal to one, y is equal to 3/2. When x is equal to two, y is equal to 3/4. And so on and so forth. And notice, because our common ratios are the reciprocal of each other, that these two graphs look like they've been flipped over, they look like they've been flipped horizontally or flipped over the y axis. They're symmetric around that y axis. And what you will see in exponential decay is that things will get smaller and smaller and smaller, but they'll never quite exactly get to zero. It'll approach zero. It'll asymptote towards the x axis as x becomes more and more positive. Just as for exponential growth, if x becomes more and more negative, we asymptote towards the x axis. So that's the introduction. I'd use a very specific example, but in general, if you have an equation of the form y is equal to A times some common ratio to the x power We could write it like that, just to make it a little bit clearer. There's a bunch of different ways that we could write it. This is going to be exponential growth, so if the absolute value of r is greater than one, then we're dealing with growth, because every time you multiply, every time you increase x, you're multiplying by more and more r's is one way to think about it. And if the absolute value of r is less than one, you're dealing with decay. You're shrinking as x increases. And I'll let you think about what happens when, what happens when r is equal to one? What are we dealing with in that situation? And it's a bit of a trick question, because it's actually quite, oh, I'll just tell you. If r is equal to one, well then, this thing right over here is always going to be equal to one and you boil down to just the constant equation, y is equal to A, so this would just be a horizontal line.

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Exponential decay intro (video) | Khan Academy

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Narrator] What we're going to do in this video is quickly review exponential growth and then use that as our platform to introduce ourselves to exponential decay. So let's review exponential growth. Let's say we have something that, and I'll do this on a table here. Just gonna make that straight.

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Exponentials Outline - MathBitsNotebook (Algebra2 - CCSS Math)

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Algebra 2 CCSS Lessons and Practice is a free site for students (and teachers) studying a second year of high school algebra.

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thinkwell burger homeschool calculus algebra chemistry video homework help tutorial tutor biology government supreme court

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Calculus, Algebra, and Chemistry videos with world-class teachers from Thinkwell.com!

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Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In this non-linear system, users are free to take whatever path through the material best serves their needs. These unique features make Virtual Nerd a viable alternative to private tutoring.