We can write a function to model exponential decay in a context. Created by Sal Khan.
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- Hello, is there is any way to determine the input value of an exponential decay function which will result in zero ?. For example, in this video, the function is [6000. (2/3)^t] this will give us the remaining viruses after t weeks. But what if we want to know after how many exact [t] weeks the virus will be eradicated. Is it possible to use logarithms in this case ?. besides I know that an exponential decay will approach zero but never become zero but there must be a certain number as input that will tell us when the value will be zero isn't? please correct me if I'm wrong thanks in advance!(6 votes)
- Technically, yes there is an input that will make that 6000(2/3)^x. However, that input is infinity.
Exponential equations have a horizontal asymptote. A horizontal asymptote is a horizontal line that a function reaches when x=infinity. A more formal way to say it is, when lim x->inf, it equals that asymptote.
For the question, the use of logarithms exhibit the same property. Rather, you'd find no solution.
Heres the example:
0=(2/3)^x // divide by 6000 on both sides
log2/3(0) = x //take the log of 2/3 on both sides.
You cannot take any log of 0. So then, it's undefined.
[Reasoning is that when you have 2/3^x = 0 (same thing as the log), no finite x value can truly equal 0 ]
hopefully that helps !!(3 votes)
- why are we multiplying by what is left after the time has passed cant we just multiply by the negative ratio that it gives us?(3 votes)
- hi! for a exponential decay since the formula is f(t)=a(1-r) its making me put "#(#)t instead of the whole formula. why is that? and fractions too like for ex 1/3 is on the question and it puts 2/3s(2 votes)
- The second question has to do with starting from 1. If you loose 1/3 of the value every year, then you are actually keeping (1-1/3)=(2/3) of the value. I saw one of the problems that had 1/4 of trees getting cut down each year. That would mean you still have (3/4) of the trees left each year. As to the first point, the formula is f(t)=a(1-r)^t, and you should be able to use the a^b button to show the exponent. If you just put t, you are multiplying it by t, not taking it to the t power.(2 votes)
- One of the questions we were given was worded somewhat vaguely and I couldn't think of how to describe what I meant in 'math-ese'. The question said that 500 new people were added to a town every year. But there are at least two ways to interpret this.
Suppose there are two towns. Abilene and Blaine. They both have the same starting population of 20,000. Abilene grows annually by adding 500 new citizens every year. Blaine also adds 500 new citizens every year, but that growth rate compounds.
So after 1 year. Abilene has 20,500. So does Blaine.
After 2 years. Abilene has 21,000. Blaine has 21,500 (adding 500+500=1000)
After 3 years Abilene has 21,500. Blaine has 23,000 (adding 500(3)=1500
So the equation for Abilene's growth would be:
20,000+500(t) where t is the number of years that pass.
Blaine's equation would be? Not 20,000+500^t. And I'm out of ideas.(1 vote)
- I think the question was asking about a town that had a growth rate like Aiblene. I'm saying this because if the growth rate was like Blaine's, more than 500 new people would be added after the first year, defying the question. I'm not very sure what the equation would be for Blaine, though. It could be t=p+500 where t is the total population and p is the population of the town the previous year. The only problem with this is that the independent variable is not the number of years that pass, but the previous population. Since the question was not very clear, I think this could be one of the right answers but I'm not very sure.
Let's see if anyone else knows if I'm right or wrong about that last part :)(2 votes)
- cant we just use linear eq.
something like v(t)=600-600(t*0.25)(1 vote)
- No, because exponential functions are about constant increase or decrease. Let's say you have $100 and you withdraw 5% every month. After you remove 5% from 100, you'd get 95. The next month, you would withdraw 5% from 95, which would get you to 90.25. And then you'd remove 5% from 90.25 the next month and so on. Thus, the only way to represent that would be y = 100 * 0.95^x. (0.95 because removing 5% from a value is the same thing as multiplying 0.95). So, if we started at 600 and consistently removed 25%, the only way to represent that would be using an exponential function.(2 votes)
- so how do you find the initial value(1 vote)
- This video should help you: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:exponential-growth-decay/x2f8bb11595b61c86:exponential-expressions/v/initial-value-and-common-ratio-of-exponential-functions
Well the first thing you need is an exponential function f(t)=at^x ( I think that's the eqation)
Then you would make x=0 then solve
substitute 0 for x
anything to the 0th power would be 1, meaning ANYTHING except 0 itself
initial value = 3
And the way you ALWAYS know the initial value is the "a" of the equation, otherwise known as the "Y-intercept"
Hope this helped :3(1 vote)
- Are there ways to solve these type questions without a table; can you solve these type questions in your mind?(1 vote)
- The thing with maths is that there are multiple methods to tackle a particular problem; it's reasonable to assume you can do this in your head(1 vote)
- [Instructor] We are told a phone sells for $600 and loses 25% of its value per year. Write a function that gives the phone's value, V of t, so value as a function of time, t years after it is sold. So pause this video, and have a go at that before we work through it together. All right, so let's just think about it a little bit. And I could even set up a table to think about what is going on. So this is t, and this is the value of our phone as a function of t. So it sells for $600. At time t equals zero, what is V of zero? Well it's going to be equal to $600. That's what it sells for at time t equals zero. Now at t equals one, what's going to happen? Well it says that the phone loses 25% of its value per year. Another way to rewrite that it loses 25% of its value per year is that it retains, it retains, 100% minus 25% of its value per year, or it retains 75% of its value per year, per year. So how much is it going to be worth after one year? Well it's going to be worth $600, $600 times 75%. Now what about year two? Well it's going to be worth what it was in year one times 75% again. So it's going to be $600 times 75% times 75%. And so you could write that as times 75% squared. And I think you see a pattern. In general, if we have gone, let's just call it t years, well then the value of our phone, if we're saying it in dollars, is just going to be $600 times, and I could write it as a decimal, 0.75, instead of 75%, to the t power. So V of t is going to be equal to 600 times 0.75 to the t power, and we're done. Let's do another example. So here, we are told that a biologist has a sample of 6,000 cells. The biologist introduces a virus that kills 1/3 of the cells every week. Write a function that gives the number of cells remaining, which would be C of t, the cells as a function of time, in the sample t weeks after the virus is introduced. So again, pause this video and see if you can figure that out. All right, so I'll set up another table again. So this is time, it's in weeks, and this is the number of cells, C. We could say it's a function of time. So time t equals zero, when zero weeks have gone by, we have 6,000 cells. That's pretty clear. Now after one week, how many cells do we have? What's C of one? Well it says that the virus kills 1/3 of the cells every week, which is another way of saying that 2/3 of the cells are able to live for the next week. And so after one week, we're going to have 6,000 times 2/3. And then after two weeks, or another week goes by, we're gonna have 2/3 of the number that we had after one week. So we're gonna have 6,000 times 2/3 times 2/3, or we could just write that as 2/3 squared. So once again, you are likely seeing the pattern here. We are going to, at time t equals zero, we have 6,000, and then we're going to multiply by 2/3 however many weeks have gone by. So the cells as a function of the weeks of t, which is in weeks, is going to be our original amount, and then however many weeks have gone by, we're going to multiply by 2/3 that many times, so times 2/3 to the t power. And we're done.