An automobile with seven years of use has a commercial value of $34057.7, but five years ago its value was $72250. If the value of such automobile varies exponentially with the time. (36-41)
36.- What was the value of the automobile when it was new?
37.- What will be the value of the automobile after 10 years of use?
38.- How many years of use will the automobile have with a value of $10277?
39.- After how long of use does the value of the automobile reduce to the half?
40.- After how many years of use the value of the automobile reduces to the fourth part of its initial value?
41.- After how many years of use does the value of the automobile reduce to the third part of its initial velocity?
the general equation would be
value = a(e)^(kt) where a is the initital vaue, and k is a constant
you end up with 2 equations:
34057.7 = a(e)^(7k) --- equation #1
72250 = a(e)^(5k) ----- eqution #2
I divided #1 by #2 to get .4713865 = e^2k
and got k = -.37604 using logs
put that back into #2 I got a = 473581
(a very unreasonable price for an automobile, but then again after 5 years it was still worth $72,000)
so now we have
Value = 473581(e)^-.37604t
and you can find any of the other answers quite easily.
To solve these problems, we can use the formula for exponential decay:
V(t) = V0 * e^(kt)
Where:
- V(t) is the value of the automobile at time t
- V0 is the initial value of the automobile
- e is the base of natural logarithms (approximately 2.71828)
- k is the decay constant
We can find the decay constant, k, using the information given in problem 36:
36. To find the initial value of the automobile (V0), we can use the formula with the given values:
$72250 = V0 * e^(7k)
To isolate V0, divide both sides of the equation by e^(7k):
V0 = $72250 / e^(7k)
37. To find the value of the automobile after 10 years (V(10)), we can use the formula:
V(10) = V0 * e^(10k)
Substitute the value of V0 from problem 36 into the formula:
V(10) = ($72250 / e^(7k)) * e^(10k)
38. To find the number of years of use with a value of $10277 (t), we can rearrange the formula:
$10277 = V0 * e^(kt)
Divide both sides of the equation by V0:
$10277 / V0 = e^(kt)
Take the natural logarithm of both sides of the equation:
ln($10277 / V0) = kt
Solve for t by dividing both sides of the equation by k:
t = ln($10277 / V0) / k
39. To find how long it takes for the value of the automobile to reduce to half of its initial value, we can use the formula:
V(t) = V0/2
Substitute V0 from problem 36 into the formula:
V(t) = ($72250 / e^(7k))/2
Set this equation equal to V0 and solve for t:
($72250 / e^(7k))/2 = V0
40. To find how many years of use it takes for the value of the automobile to reduce to a fourth of its initial value, we can use the formula:
V(t) = V0/4
Substitute V0 from problem 36 into the formula:
V(t) = ($72250 / e^(7k))/4
Set this equation equal to V0 and solve for t:
($72250 / e^(7k))/4 = V0
41. To find how many years of use it takes for the value of the automobile to reduce to a third of its initial value, we can use the formula:
V(t) = V0/3
Substitute V0 from problem 36 into the formula:
V(t) = ($72250 / e^(7k))/3
Set this equation equal to V0 and solve for t:
($72250 / e^(7k))/3 = V0
You can solve these equations using a scientific calculator or software capable of evaluating exponential functions and natural logarithms.