## Solving With Exponentials

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richardibarnes
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### Solving With Exponentials

Apologies if this is in the wrong place, I am new to the forum.

I have been stuck on this problem for weeks now and came across this site via google. Problem is as follows;

John would like to buy a new car for $10,000. His current car is worth$6,000 and will depreciate to 85% of it's value each year. John can save $350 per month towards the purchase. When is the soonest he can buy the new car? I figured I could express the problem as follows with y being the number of months (using x as multiplication sign) 6000 x (0.85/12)^y + 350y = 10000 so the number of months of saving equals the number of months for depreciation. The problem is a) is this correct and then b) I can't seem to work it where the exponential (y) is the same as a mulitplier on the same side. Any help (or pointing me where this should be in the forum) would be much appreciated! anonmeans Posts: 84 Joined: Sat Jan 24, 2009 7:18 pm Contact: ### Re: Solving With Exponentials John would like to buy a new car for$10,000. His current car is worth $6,000 and will depreciate to 85% of it's value each year. This doesn't make sense. If its value is$6K and it "will depreciate to 85% of its value", then it will depreciate once to $5.1K. I think they probably meant to say that, each year, it will depreciate to 85% of the previous year's value, or "by" 85%. Then the problem makes sense. John can save$350 per month towards the purchase. When is the soonest he can buy the new car?
So he's going to use his car as a trade-in, and he's saving up so he can add to the trade-in value. But the longer he waits and saves, the less he'll get for the trade-in.
I figured I could express the problem as follows with y being the number of months (using x as multiplication sign)

6000 x (0.85/12)^y + 350y = 10000
FYI: Use "*" for "times", so there isn't confusion with the variable. (Thanks for saying what you meant, though. That's very helpful!)
so the number of months of saving equals the number of months for depreciation. The problem is a) is this correct and then b) I can't seem to work it where the exponential (y) is the same as a mulitplier on the same side.
The set-up looks like you have the right idea, using monthly-compounded "decay" for the value of the car, and linear growth for the value of his savings. But you're right about the solving problem. Because part is linear and part is exponential, this is going to be a problem to solve algebraically. You may have to solve numerically.

Edit: Wolfram Alpha comes up with something really nasty (this), so you're definitely right about the solving problem. What class assigned you this problem? What did the class cover just before assigning this? If you're supposed to use numerical methods, you can edit the equation to be like this:

$6,000\, \left(1\, -\, \dfrac{0.16142}{12}\right)^y\, +\, 350y\, -\, 10,000\, =\, 0$

and then find the solution to this by using what they show here.

I used the regular compound-growth formula, because yours didn't return the right value for 12 months of negative growth:

$6,000\, \left(\dfrac{0.85}{12}\right)^{12}\, \approx \, 9.572\, \times\, 10^{-11}$

which is obviously very wrong. Using the regular form for compounded whatever (and using the interest rate that gives the right answer for the end of the year) gives you:

$6,000\, \left(1\, -\, \dfrac{0.16142}{12}\right)^{12}\, \approx\, 5,100.01739$

Otherwise, just use the the yearly form and linear growth:

$6,000\, (0.85\, t)\, +\, 350\, (12t)\, =\, 10,000$

Then convert whatever fractional form you get to "months and years" or just "months" by multiplying by 12.