The table below gives the value of a car over two years. Using an exponential best-fit model, after how many years is the car valued at 25% of the original value?
Year 0 0.5 1.0 1.5 2.0
Value ($) 15,500.00 14,560.00 13,911.20 12,969.66 12,204.09
correct clara
To find out after how many years the car is valued at 25% of the original value, we will use an exponential best-fit model equation to estimate the car's value at different time points.
The exponential best-fit model can be represented by the equation:
V(t) = V₀ * e^(kt)
Where:
V(t) is the value of the car at time t,
V₀ is the initial value of the car,
k is the growth/decay constant,
t is the time in years,
e is the base of the natural logarithm (approximately 2.71828).
To find the value of k, we need to take the natural logarithm of both sides of the equation and solve for k:
ln(V(t)) = ln(V₀ * e^(kt))
ln(V(t)) = ln(V₀) + ln(e^(kt))
ln(V(t)) = ln(V₀) + kt * ln(e)
ln(V(t)) - ln(V₀) = kt
Now, let's apply this equation to our problem:
V₀ = 15,500 (initial value of the car)
t = 2 (given time in years)
V(t) = 12,204.09 (value of the car after 2 years)
Plugging in the values, we have:
ln(12,204.09) - ln(15,500) = k * 2
Now, we can solve for k:
ln(12,204.09/15,500) = -2k/2
ln(0.787834) = -k
Now, we need to find the time when the car is valued at 25% of the original value (V₀).
Let's set up the equation:
0.25V₀ = V₀ * e^(kt)
Simplifying the equation:
0.25 = e^(kt)
Taking the natural logarithm of both sides:
ln(0.25) = ln(e^(kt))
Using the properties of logarithms:
ln(0.25) = kt
Now, let's solve for t:
-1.3862944 = -k * t
Dividing both sides by -k:
t = -1.3862944 / -k
Substituting the value of k we found earlier:
t = -1.3862944 / -(ln(0.787834))
Evaluating the expression:
t ≈ 1.08 years
Therefore, after approximately 1.08 years, the car will be valued at 25% of the original value.