To find the correct exponential equation that fits the given points (0, 30000) and (1, 22500), we can use the general form of an exponential decay equation, which is:
\[ V(t) = V_0 \cdot e^{-kt} \]
Where:
- \( V(t) \) is the value of the car at time \( t \) (in years).
- \( V_0 \) is the initial value of the car (value at \( t = 0 \)).
- \( k \) is a decay constant.
- \( e \) is the base of the natural logarithm.
- \( t \) is the number of years since purchase.
Given the point (0, 30000), we start with:
- \( V(0) = 30000 \)
This implies that: \[ V_0 = 30000 \]
Next, using the point (1, 22500), we have: \[ V(1) = 30000 \cdot e^{-k} = 22500 \]
To solve for \( k \), we can divide both sides by 30000: \[ e^{-k} = \frac{22500}{30000} = 0.75 \]
Now, we take the natural logarithm of both sides: \[ -k = \ln(0.75) \]
Therefore: \[ k = -\ln(0.75) \]
Now we can express the equation: \[ V(t) = 30000 \cdot e^{-\ln(0.75) \cdot t} \]
This can also be expressed in a simpler base form using the property of logarithms, since: \[ e^{-\ln(0.75)} = \frac{1}{0.75} \]
Thus, in terms of decay: \[ V(t) = 30000 \cdot (0.75)^t \]
So the full exponential equation that represents the value of the car since purchase is:
\[ V(t) = 30000 \cdot (0.75)^t \]
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