7
Nick bought a new car.
Each year the car depreciates in value by 12%.
Work out the number of years it takes for the car to half in value.
(Total for question 7 is 3 mark
0.0008 W
SIZE 1
THAILAND



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.