Answer :
Sure, let's work through this problem step-by-step.
### Step-by-Step Solution:
Given:
- The car depreciates by 12% each year.
- We need to find out how many years it will take for the car to half in value.
We can use the formula for exponential decay to solve this problem:
[tex]\[ V(t) = V_0 \cdot (1 - r)^t \][/tex]
Where:
- [tex]\( V(t) \)[/tex] is the value of the car after [tex]\( t \)[/tex] years.
- [tex]\( V_0 \)[/tex] is the initial value of the car.
- [tex]\( r \)[/tex] is the depreciation rate per year.
- [tex]\( t \)[/tex] is the number of years.
Since we want to find when the car's value is half of the initial value:
[tex]\[ \frac{1}{2} V_0 = V_0 \cdot (1 - r)^t \][/tex]
The initial value [tex]\( V_0 \)[/tex] can cancel out from both sides:
[tex]\[ \frac{1}{2} = (1 - r)^t \][/tex]
Now, plug in the given depreciation rate [tex]\( r = 0.12 \)[/tex]:
[tex]\[ \frac{1}{2} = (1 - 0.12)^t \][/tex]
Simplify the decay factor:
[tex]\[ \frac{1}{2} = 0.88^t \][/tex]
To solve for [tex]\( t \)[/tex], we need to take the logarithm of both sides. Using the natural logarithm (ln) will be helpful here:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(0.88^t) \][/tex]
Using the power rule of logarithms ([tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]):
[tex]\[ \ln\left(\frac{1}{2}\right) = t \cdot \ln(0.88) \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{1}{2}\right)}{\ln(0.88)} \][/tex]
Calculate the natural logarithms:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(0.5) \approx -0.6931 \][/tex]
[tex]\[ \ln(0.88) \approx -0.1278 \][/tex]
Substitute these values back into the equation:
[tex]\[ t = \frac{-0.6931}{-0.1278} \approx 5.42 \][/tex]
So, it takes approximately 5.42 years for the car to depreciate to half its initial value.
### Final Answer:
It takes approximately 5.42 years for the car to depreciate to half its initial value.
### Step-by-Step Solution:
Given:
- The car depreciates by 12% each year.
- We need to find out how many years it will take for the car to half in value.
We can use the formula for exponential decay to solve this problem:
[tex]\[ V(t) = V_0 \cdot (1 - r)^t \][/tex]
Where:
- [tex]\( V(t) \)[/tex] is the value of the car after [tex]\( t \)[/tex] years.
- [tex]\( V_0 \)[/tex] is the initial value of the car.
- [tex]\( r \)[/tex] is the depreciation rate per year.
- [tex]\( t \)[/tex] is the number of years.
Since we want to find when the car's value is half of the initial value:
[tex]\[ \frac{1}{2} V_0 = V_0 \cdot (1 - r)^t \][/tex]
The initial value [tex]\( V_0 \)[/tex] can cancel out from both sides:
[tex]\[ \frac{1}{2} = (1 - r)^t \][/tex]
Now, plug in the given depreciation rate [tex]\( r = 0.12 \)[/tex]:
[tex]\[ \frac{1}{2} = (1 - 0.12)^t \][/tex]
Simplify the decay factor:
[tex]\[ \frac{1}{2} = 0.88^t \][/tex]
To solve for [tex]\( t \)[/tex], we need to take the logarithm of both sides. Using the natural logarithm (ln) will be helpful here:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(0.88^t) \][/tex]
Using the power rule of logarithms ([tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]):
[tex]\[ \ln\left(\frac{1}{2}\right) = t \cdot \ln(0.88) \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{1}{2}\right)}{\ln(0.88)} \][/tex]
Calculate the natural logarithms:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(0.5) \approx -0.6931 \][/tex]
[tex]\[ \ln(0.88) \approx -0.1278 \][/tex]
Substitute these values back into the equation:
[tex]\[ t = \frac{-0.6931}{-0.1278} \approx 5.42 \][/tex]
So, it takes approximately 5.42 years for the car to depreciate to half its initial value.
### Final Answer:
It takes approximately 5.42 years for the car to depreciate to half its initial value.
