Algebra 2

OK.....I'll take a stab at this one.

We have, after one year

V = 25000e^(T*N)

Where "V" is the value of the car after T years. We need to find "N"

So, we have (after 1 year)

21250 = 2500e^(1*N)

Divide both sides by 25000

(21250/25000) = e^(1*N)

Now, take the LN of both sides

LN(21250/25000) = LN e^(1*N)

By the property pf logs, the LN e^(1*N) just becomes (1 * N) or just N !!

So we have

LN(21250/25000) = N

Since this is just a messy decimal, let's leave N as it is!!

So, our function is

V = 25000e^(T * LN(21250/25000))

To find the value after 5 years we have

squidhardy:

The initail value of a car is $25,000. After one year, the value of the car is $21,250. Write an exponential function to model the expected value of the car. Estimate the value of the car after 5 years.

Hi Squidhardy and CPhill
there is a formula but I want to show you the logic behind it.
Say the deflation rate is r per year. This will mean the following

Original value = $25000
After 1 yr = $25000 * (1-r)
After 2 yr = $25000 * (1-r) ²
After 3 yr = $25000 * (1-r) ³
After 4 yr = $25000 * (1-r) ⁴
After 5 yr = $25000 * (1-r) ⁵
After n years = $2500 * (1-r) ⁿ

the general formula is F _n=P(1-r) ⁿ
Where F is future value and P is the principal value.

so we have
F _n = $25000 * (1-r) ⁿ
and
$25000 * (1-r) = $21,250 [Value after 1 year is used to find r]
1-r = 21250/25000
1-21250/25000=r

Sorry Chris,
I just realised you answer is the same as mine. I had better look at what you did a little more thoroughly.
(I think my way is a little easier to understand)