Writing functions with exponential decay | Algebra 1 | Khan Academy

We are told a phone sells for six hundred dollars and loses 25% of its value per year. Write a function that gives the phone's value ( v(t) ) so value is a function of time ( t ) years after it is sold. So pause this video and have a go of that before we work through it together.

All right, so let's just think about it a little bit. I could even just 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 six hundred dollars. So time ( t = 0 ), what is ( v(0) )? Well, it's going to be equal to six hundred dollars. That's what it sells for time ( t = 0 ).

Now, ( t = 1 ), what's going to happen? Well, it says that the phone loses 25% of its value per year. Another way to rewrite it is that it loses 25% of its value per year is that it retains 100% minus 25% of its value per year, or it retains 75% of its value per year. So how much is it going to be worth after one year? Well, it's going to be worth ( 600 \times 0.75 ).

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 0.75 \times 0.75 ) and so you could write that as times ( 0.75^2 ). 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 0.75^t ). So ( v(t) ) is going to be equal to ( 600 \times 0.75^t ), 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 one third of the cells every week. Write a function that gives the number of cells remaining, which would be ( c(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 = 0 ), when zero weeks have gone by, we have six thousand cells. That's pretty clear. Now after one week, how many cells do we have? What's ( c(1) )? Well, it says that the virus kills one-third of the cells every week, which is another way of saying that two-thirds of the cells are able to live for the next week.

So after one week we're going to have ( 6000 \times \frac{2}{3} ). Then after two weeks, or another week goes by, we're gonna have two-thirds of the number that we had after one week. So we're gonna have ( 6000 \times \frac{2}{3} \times \frac{2}{3} ) or we could just write that as ( \left(\frac{2}{3}\right)^{2} ).

Once again, you are likely seeing the pattern here. We are going to at time ( t = 0 ) we have six thousand, and then we're going to multiply by two-thirds however many times, however many weeks have gone by. So the cells as a function of the weeks ( 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 ( \left(\frac{2}{3}\right)^{t} ) and we're done.