7 nick bought a new car. each year the car depreciates in value by 12%. work out the number of years it…

Explanation:

Step1: Set up the depreciation formula for Nick's car

Let the initial value of the car be $V_0$ and the value after $t$ years be $V$. The depreciation formula is $V = V_0(1 - r)^t$, where $r = 0.12$. We want to find $t$ when $V=\frac{1}{2}V_0$. So, $\frac{1}{2}V_0=V_0(1 - 0.12)^t$. Canceling out $V_0$ (since $V_0\neq0$), we get $\frac{1}{2}=(0.88)^t$.

Step2: Take the natural - logarithm of both sides

$\ln(\frac{1}{2})=\ln(0.88^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(\frac{1}{2}) = t\ln(0.88)$.

Step3: Solve for $t$

$t=\frac{\ln(\frac{1}{2})}{\ln(0.88)}=\frac{-\ln(2)}{\ln(0.88)}\approx\frac{- 0.6931}{-0.1278}\approx5.43$.

Step4: Set up the compound - interest formula for Fearne's investment

The compound - interest formula is $A = P(1 + r)^n$, where $P = 5600$, $r=0.02$, and $A = 6061.62$. So, $6061.62=5600(1 + 0.02)^n$.

Step5: Divide both sides by 5600

$\frac{6061.62}{5600}=(1.02)^n$. $1.082432=(1.02)^n$.

Step6: Take the natural - logarithm of both sides

$\ln(1.082432)=\ln(1.02^n)$. Using the property $\ln(a^b)=b\ln(a)$, we get $\ln(1.082432)=n\ln(1.02)$.

Step7: Solve for $n$

$n=\frac{\ln(1.082432)}{\ln(1.02)}\approx\frac{0.0791}{0.0198}\approx4$.

Answer:

For Nick's car: approximately 5.43 years For Fearne's investment: 4 years