I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.
With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get:
Therefore your result will be too large.
If you use Chris L T 521's formula:
then where p is the percentage of decrease.
Now compare:
decrease by 2% ====> ln(0.98) = -0.020202...
decrease by 5% ====> ln(0.95) = -0.051293...
decrease by 10% ====> ln(0.9) = -0.10536...
...
decrease by 20% ====> ln(0.8) = -0.22314...
You see that the approximation of k = -p is only sufficient for p < 10%.
Seeing the way that Chris approached the problem,
I am obligated to ask: Is someone teaching that the -formula
. . is the only way to handle depreciation?
If so, they are doing a great disservice.
While is highly applicable to decay and population problems,
I feel that it requires more work when applied to depreciation problem,
. . especially when they give us the depreciation rate.
Consider the original problem . . .
. . Initial value: $20,000
. . Depreciation rate: 30% (annually)
We start with: .
When
. . Hence, the function (so far) is: .
When
. .
Hence, the function is: .
This equation is correct . . . I have no complaint.
But simplify the equation: .
Since , we have: .
We had this equation way back then, just "eyeballing" the problem.
The difficulty lies in the word "always" ...
With the given problem the variable t is a discret number, that means actually you are dealing with a geometric sequence:
If a value is increasing or decreasing continuously your method can't be applied, then you have to use the equation which was posted by ChrisLT521