question 3\na lump sum of $13,500 was deposited into an account earning 9.6% annual interest, compounded…

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. Given $P = 13500$, $A=25000$, $r = 0.096$ (since $9.6%=0.096$), and $n = 4$ (compounded quarterly). Substitute these values into the formula: $25000=13500(1 +\frac{0.096}{4})^{4t}$.

Step2: Simplify the equation

First, simplify $1+\frac{0.096}{4}=1 + 0.024=1.024$. The equation becomes $\frac{25000}{13500}=(1.024)^{4t}$. $\frac{50}{27}=(1.024)^{4t}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{50}{27})=\ln((1.024)^{4t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{50}{27}) = 4t\ln(1.024)$. We know that $\ln(\frac{50}{27})\approx\ln(50)-\ln(27)\approx3.912 - 3.296=0.616$ and $\ln(1.024)\approx0.0237$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{50}{27})}{4\ln(1.024)}$. $4\ln(1.024)\approx4\times0.0237 = 0.0948$. $t=\frac{0.616}{0.0948}\approx6.5$.

Answer:

$6.5$