Question

1. francis buys a car for $30 000. he estimates that the value of the car will decrease by 20% annually. after how many years, will the value of the car be $7034?
2. after how many years will a capital double when placed at a simple interest rate of 4%?
3. after how many years, rounded to the nearest tenth, will a capital double when placed at an interest rate of 4% compounded annually?

Explanation:

Problem 3

Step1: Define depreciation formula

The car's value follows exponential depreciation: $V(t) = V_0(1 - r)^t$, where $V_0=\$30000$, $r=0.2$, $V(t)=\$7034$.

Step2: Substitute known values

$7034 = 30000(1 - 0.2)^t$

Step3: Isolate the exponential term

$\frac{7034}{30000} = 0.8^t$
$0.23447 \approx 0.8^t$

Step4: Solve for t using logarithms

Take natural log of both sides: $\ln(0.23447) = t\ln(0.8)$
$t = \frac{\ln(0.23447)}{\ln(0.8)}$

Step5: Calculate the value

$t \approx \frac{-1.459}{-0.2231} \approx 6.5$

Step1: Define simple interest formula

Simple interest total value: $A = P(1 + rt)$. For doubling, $A=2P$.

Step2: Substitute A=2P and r=0.04

$2P = P(1 + 0.04t)$

Step3: Cancel P and isolate t

$2 = 1 + 0.04t$
$0.04t = 1$

Step4: Solve for t

$t = \frac{1}{0.04}$

Step1: Define compound interest formula

Annual compounding: $A = P(1 + r)^t$. For doubling, $A=2P$.

Step2: Substitute A=2P and r=0.04

$2P = P(1 + 0.04)^t$

Step3: Cancel P and isolate exponential term

$2 = 1.04^t$

Step4: Solve for t using logarithms

Take natural log of both sides: $\ln(2) = t\ln(1.04)$
$t = \frac{\ln(2)}{\ln(1.04)}$

Step5: Calculate and round

$t \approx \frac{0.6931}{0.03922} \approx 17.7$

Answer:

6.5 years

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Problem 4