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:
\(\[ V(t) = V_0 \cdot (1 - r)^t \]\)
Where:
- \(\( V(t) \)\) is the value of the car after \(\( t \)\) years.
- \(\( V_0 \)\) is the initial value of the car.
- \(\( r \)\) is the depreciation rate per year.
- \(\( t \)\) is the number of years.
Since we want to find when the car's value is half of the initial value:
\(\[ \frac{1}{2} V_0 = V_0 \cdot (1 - r)^t \]\)
The initial value \(\( V_0 \)\) can cancel out from both sides:
\(\[ \frac{1}{2} = (1 - r)^t \]\)
Now, plug in the given depreciation rate \(\( r = 0.12 \)\):
\(\[ \frac{1}{2} = (1 - 0.12)^t \]\)
Simplify the decay factor:
\(\[ \frac{1}{2} = 0.88^t \]\)
To solve for \(\( t \)\), we need to take the logarithm of both sides. Using the natural logarithm (ln) will be helpful here:
\(\[ \ln\left(\frac{1}{2}\right) = \ln(0.88^t) \]\)
Using the power rule of logarithms (\(\( \ln(a^b) = b \cdot \ln(a) \)\)):
\(\[ \ln\left(\frac{1}{2}\right) = t \cdot \ln(0.88) \]\)
Solve for \(\( t \)\):
\(\[ t = \frac{\ln\left(\frac{1}{2}\right)}{\ln(0.88)} \]\)
Calculate the natural logarithms:
\(\[ \ln\left(\frac{1}{2}\right) = \ln(0.5) \approx -0.6931 \]\)
\(\[ \ln(0.88) \approx -0.1278 \]\)
Substitute these values back into the equation:
\(\[ t = \frac{-0.6931}{-0.1278} \approx 5.42 \]\)
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.
