Answer :
To determine the annual percentage decrease in value [tex]\(x\)[/tex] given that a car's value halves in 6 years, we can use a mathematical approach to solve this problem.
1. Understand the Problem:
- Let's denote the initial value of the car as [tex]\(V\)[/tex].
- After 6 years, the value of the car is [tex]\(V/2\)[/tex].
- The car's value decreases by [tex]\(x\)[/tex]% each year.
2. Write the Formula for Depreciation:
The formula for the value of the car after [tex]\(t\)[/tex] years, considering an annual decrease of [tex]\(x\% ,\)[/tex] is:
[tex]\[ \text{final value} = \text{initial value} \times (1 - \frac{x}{100})^t \][/tex]
3. Apply the Given Information:
After 6 years, the final value is [tex]\(V/2\)[/tex]. So,
[tex]\[ \frac{V}{2} = V \times (1 - \frac{x}{100})^6 \][/tex]
4. Simplify the Equation:
Divide both sides by [tex]\(V\)[/tex]:
[tex]\[ \frac{1}{2} = (1 - \frac{x}{100})^6 \][/tex]
5. Solve for [tex]\((1 - \frac{x}{100})\)[/tex]:
Take the natural logarithm (ln) of both sides to solve for the base term:
[tex]\[ \ln(\frac{1}{2}) = \ln((1 - \frac{x}{100})^6) \][/tex]
[tex]\[ \ln(\frac{1}{2}) = 6 \cdot \ln(1 - \frac{x}{100}) \][/tex]
6. Isolate the Logarithmic Term:
[tex]\[ \ln(1 - \frac{x}{100}) = \frac{\ln(\frac{1}{2})}{6} \][/tex]
7. Evaluate the Terms:
The natural logarithm of 1/2 is a known value, [tex]\(\ln(\frac{1}{2}) = -0.6931\)[/tex]. Plug in this value:
[tex]\[ \ln(1 - \frac{x}{100}) = \frac{-0.6931}{6} \][/tex]
[tex]\[ \ln(1 - \frac{x}{100}) \approx -0.1155 \][/tex]
8. Exponentiate Both Sides to Remove Logarithm:
[tex]\[ 1 - \frac{x}{100} = e^{-0.1155} \][/tex]
[tex]\[ 1 - \frac{x}{100} \approx 0.8900 \][/tex]
9. Solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{x}{100} = 1 - 0.8900 \][/tex]
[tex]\[ \frac{x}{100} \approx 0.1100 \][/tex]
[tex]\[ x \approx 11.0 \][/tex]
10. Finalize the Answer and Round to 1 Decimal Place:
[tex]\[ x \approx 10.9 \][/tex]
Therefore, the annual percentage decrease [tex]\(x\)[/tex] is approximately 10.9%.
1. Understand the Problem:
- Let's denote the initial value of the car as [tex]\(V\)[/tex].
- After 6 years, the value of the car is [tex]\(V/2\)[/tex].
- The car's value decreases by [tex]\(x\)[/tex]% each year.
2. Write the Formula for Depreciation:
The formula for the value of the car after [tex]\(t\)[/tex] years, considering an annual decrease of [tex]\(x\% ,\)[/tex] is:
[tex]\[ \text{final value} = \text{initial value} \times (1 - \frac{x}{100})^t \][/tex]
3. Apply the Given Information:
After 6 years, the final value is [tex]\(V/2\)[/tex]. So,
[tex]\[ \frac{V}{2} = V \times (1 - \frac{x}{100})^6 \][/tex]
4. Simplify the Equation:
Divide both sides by [tex]\(V\)[/tex]:
[tex]\[ \frac{1}{2} = (1 - \frac{x}{100})^6 \][/tex]
5. Solve for [tex]\((1 - \frac{x}{100})\)[/tex]:
Take the natural logarithm (ln) of both sides to solve for the base term:
[tex]\[ \ln(\frac{1}{2}) = \ln((1 - \frac{x}{100})^6) \][/tex]
[tex]\[ \ln(\frac{1}{2}) = 6 \cdot \ln(1 - \frac{x}{100}) \][/tex]
6. Isolate the Logarithmic Term:
[tex]\[ \ln(1 - \frac{x}{100}) = \frac{\ln(\frac{1}{2})}{6} \][/tex]
7. Evaluate the Terms:
The natural logarithm of 1/2 is a known value, [tex]\(\ln(\frac{1}{2}) = -0.6931\)[/tex]. Plug in this value:
[tex]\[ \ln(1 - \frac{x}{100}) = \frac{-0.6931}{6} \][/tex]
[tex]\[ \ln(1 - \frac{x}{100}) \approx -0.1155 \][/tex]
8. Exponentiate Both Sides to Remove Logarithm:
[tex]\[ 1 - \frac{x}{100} = e^{-0.1155} \][/tex]
[tex]\[ 1 - \frac{x}{100} \approx 0.8900 \][/tex]
9. Solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{x}{100} = 1 - 0.8900 \][/tex]
[tex]\[ \frac{x}{100} \approx 0.1100 \][/tex]
[tex]\[ x \approx 11.0 \][/tex]
10. Finalize the Answer and Round to 1 Decimal Place:
[tex]\[ x \approx 10.9 \][/tex]
Therefore, the annual percentage decrease [tex]\(x\)[/tex] is approximately 10.9%.
