Exponential equation word problem
Sal models a context that concerns a bank savings account. The model turns out to be an exponential equation. Created by Sal Khan.
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- Why does sal say 1.2 after minute 1? I get him saying .2 as that comes from 20% but why 1.2?(17 votes)
- This is because he is just adding the 20 percent change to the original hundred percent to find out the answer. He is finding 20 percent of 6250 and then adding it to 6250 (which would be 100 percent) which becomes 120 percent or 1.2. This is the best explanation I can give, sorry if you didn’t understand. If you didn’t get this you should go and check out the previous units, where he explains it better.(2 votes)
- Why (1.2)^r and not (1.2)^(r-1) ?(5 votes)
- I'm assuming you mean "t" when you say "r". The reason it's not (1.2)^(t-1) is because 6250 is multiplied by 1.2 at the end of EACH year, starting with the FIRST.(5 votes)
- I tried doing this utilizing the geometric series equation (Sn = a(1-r^n)/1-r, but I don't seem to get the same answer(6 votes)
- why when you multiply the percent, why wouldn't it just be (.20)? Why add a one? That's whats just like throwing me off. Like where is this 1 coming from?(3 votes)
- If you multiply it by 0.2 you end up with 1/5th of what you began with. It's supposed to increase. The "1" represents what you started with that year, and the additional 0.2 represents the 20% increase that year. Because 0.2 is 20% of one.(3 votes)
- How do you identify if you have an exponential sequence?(2 votes)
- You know you have an exponential sequence if the ratio between consecutive terms is a constant. What does this mean? Let's do an example:
Here is a sequence:
1, 2, 2, 4, 8, 32, 256
Is it exponential? Find the ratios
2/1 = 2
2/2 = 1
4/2 = 2
8/4 = 2
32/8 = 4
256/32 = 8
The ratios are all two for the first couple terms, but then increase to 4 and 8, so this can't be exponential.
1, 1.1, 1.21, 1.331, 1.4641, 1.61051
Is it exponential? Find the ratios.
1.1/1 = 1.1
1.21/1.1 = 1.1
1.331/1.21 = 1.1
1.4641/1.331 = 1.1
1.61051/1.4641 = 1.1
All the ratios are the same, so the sequence must be exponential.
Basically, divide a term in a sequence by the previous term. If the quotient is always the same number, then you have an exponential sequence.
(Note: the quotient could be fractional and even negative. 1, -0.5, 0.25. -0.125, 0.0625. Here, the quotient is always -0.5)
Just for fun, can you tell me what the rule is for the first sequence I gave?(5 votes)
- Is this the fastest way:
6250(6/5)^t = 12960
(6^t)/(5^t) = 1296/625 = (6^4)/(5^4)
(6/5)^t = (6/5)^4
- That could be the fastest way for many people, and it's extra nice because you don't necessarily need a calculator to solve it. Sal's way works best for me (since I use a calculator), but any way is fine if it gives you the correct answer.(3 votes)
- why does my calculator give log 12960 as 4.11261, whereas Sal calculates it as 2.0736? I know log to the base 10 would give us log 10,000 is 4.....(2 votes)
- The 2.0736 comes from 12960/6250. The total money in the account after t years can be modeled by A=6250*1.2^t. We want to know how long it will take for the account to reach 12960. Therefore, we plug in the 12960 into the equation: (12960=6250*1.2^t) After dividing and isolating t we are left with: (2.0736=1.2^t) This is where Sal uses the log function to solve for t.(2 votes)
- At7:24why does it not matter what base logarithm you use?(2 votes)
I think it is because of the change of base property of logs. See this video:
- I need help on this question:
- The starting equation is:
Since 4=2^2, we will add it in
We multiply the 2 and x-3 since it is a power to a power.
Then we add 2x+6 and x+1.
From here we subtract 8 from both sides
If f(x)<g(x) then In(f(x))<In(g(x))
Then we apply the log rule:
I don't know what to do from here, so I just plugged it in a calculator, and I got...
And we are done.(2 votes)
- Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half.
The half-life of the isotope dubnium-263 is 29 seconds. A sample of dubnium-263 was first measured to have 1024 atoms. After t seconds, there were only 32 atoms of this isotope remaining.
Write an equation in terms of t that models the situation.
Here is my answer:
32 = 1024(1/2)^(t/29)
Please verify my answer..Provide correction if I have something wrong..(2 votes)
- Yep, you got it. Is there something in particular you are having trouble with that you're asking about?(1 vote)
Liam opened a savings account and put $6,250 in it. Each year, the account increases by 20%. How many years will it take the account to reach $12,960? Write an equation that models the situation. Use t to represent the number of years since Liam opened the account. So I encourage you to pause this video and actually try to do it on your own first. Try to write this equation that models the situation using the variable t in the way they described. And then actually answer the question, how many years will it take for the account to reach $12,960? Well, let's just think about it. So t is to represent the number of years since Liam opened the account. So let's just say it's been 0 years since Liam opened the account. How much is he going to have? Well, he's just going to have $6,250 in it. That's how much he starts with. Now, let's say it's been one year since he opened the account. How much will he have? Well, he's going to have $6,250 times-- or let's write it this way-- plus 20% 6,250. It grows 20% every year. So this is how much he started the year with, and then he gets another 20% of that 6,250. If we factor out a 6,250, this is equal to 6,250 times 1 plus 20%, or we could write that as 0.2. Which is equal to 6,250 times 1.2. Now, how much is he going to have at the end of two years? Well, he's going to have the same amount that he still had at the end of one year, times 1.2. Because it grew by 20% again. So he's going to have the amount that he had at the end of one year times 1.2, which is equal to 6,250 times 1.2, times 1.2. Which is equal to 6,250 times 1.2 squared. I think you might see where this is going. Or I could even write it like this, order of operations, you do the exponent first. So what about after three years? So after three years, well, we're just going to compound. We're going to multiply by 1.2 once again. So then he's going to have 6,250 times 1.2 to the third power. And so after t years, we're going to multiply by 1.2 that many times. So after t years in his account, he's going to have 6,250 times 1.2 to the t'th power. 1.2 to the t'th power, or to the t power. I don't want get confused with the thing that you use to take bites with. Anyway. So they say, write an equation that models the situation. So we want to figure out how many years will it take the account to reach 12,960? So we essentially want to say, when is the account going to be $12,960? Or we could write 12,960, when is that going to be equal to 6,250 times 1.2 to the t power? So that's the equation right over there that models the situation. And then we need to think about how we can actually go about solving this thing. Well, a natural thing is to isolate the t variable. Let's divide both sides by 6,250. So we could get-- and if we flip the two sides, we could get 1.2 to the t power is equal to-- well let me write this, 12,960 divided by 6,250. And since they're both divisible by 10, why don't we divide them both by 10? So it's 1,296 divided by 625. And there's several ways that you could solve this problem at this point. One way, if you feel confident that this is going to have an integer answer right over here, you could literally just try to use your calculator and multiply 1.2 enough times to get whatever number this is. And so we could do it that way. And as we'll see, there's a more systematic way of doing it once you learn about logarithms, and I'll do that at the end. But I'll do that last just in case you haven't been exposed to logarithms yet. So you could literally say-- so let me just exit out of everything. So you could literally say OK, let's see, 1,296 divided by 625 is this value. So let's see how many times we have to multiply by 1.2. 1.2 times 1.2 gets us-- well, that doesn't get us close enough. So let's try it three times. So let's take that same number. Let's just take 1.2, let's raise it. Let's raise 1.2. Let's just do it three times. Times 1.2 times 1.2. That still doesn't get us there. What if we were to multiply by 1.2 one more time? Well, that actually gets us there. And we just did this by brute force. 1.2 to the fourth power will give us this value. So that's one way, kind of a brute-force way, of figuring out that t is equal to 4. Another way, this just might be a little bit less intuitive, it might jump out of it, gee, this looks like some type of a power of 5. We know that 5 to the first is 5, 5 squared is 25, 5 to the third is 125, 5 to the fourth is 625. And so you might recognize this right over here is 5 to the fourth. And it's actually a little bit harder to recognize that this right over here is 6 to the fourth power. And this right over here is 6/5. So we could rewrite this as 6/5 to the t is equal to 6 to the fourth over 5 to the fourth. Which is the same thing as 6/5 to the fourth power. So here you'd say, well, 6/5 to the t needs to be equal to 6/5 to the fourth power. t must be equal to 4. Now, this is nice when you can recognize that this is something raised to the fourth power, which isn't easy to do. Or if you know this is an integer, and you can just keep multiplying 1.2-- if you know it's a low integer. But the systematic way of doing it is to actually use logarithms. And there's many videos on Khan Academy about how to use logarithms. But if you're more concerned with, well gee, if I just want to figure out 1.2 to what power is equal to this thing, what you would do-- and we prove this in other videos-- is if you say, look, let's take the thing that we want 1.2 to some power to be. Let's take the logarithm of that. And actually, you could take a logarithm any base. Your calculator tends to have a natural log, which is base e, and a log base 10. We could just take a log base 10. So let's do that. So we'll take the logarithm of what we want to get to, 2.0736, and divide that by the thing that we're trying to take the power of to get to this number. So divided by the logarithm of 1.2. And once again, we prove this-- actually, I wanted to divide. So let me insert division symbol. So once again, it might look like a little voodoo right here. We prove it in other videos, but if you wanted to use a calculator to calculate things like this, because sometimes it won't be a nice integer number of years. It might be 3 and 1/2 years, or it might be 7.1234 years, whatever it might be. This will give you a more precise answer. So what do you want to get to? You want to get to 2.0736. What are you raising to some power? 1.2. Divide the log of the thing you're trying to get to divided by the log of what the base that you're trying to raise to a power, and you click Enter. And then you get-- so this is literally another way of saying that 1.2 to the fourth power is going to be 2.0736. So once again, if this looks like voodoo, you don't know what logarithms are, we have videos on Khan Academy on that. But there's multiple ways to tackle it, especially this problem where the answer was a little bit simpler.