question 16\na lump - sum of $100,000 was deposited into an account earning 8.0% 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), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. Given $P = 100000$, $A=320000$, $r = 0.08$ (since $8%=0.08$), and $n = 12$ (compounded monthly).

Step2: Substitute values into the formula

$320000=100000(1 +\frac{0.08}{12})^{12t}$. First, divide both sides by 100000: $\frac{320000}{100000}=(1+\frac{0.08}{12})^{12t}$, so $3.2=(1+\frac{0.08}{12})^{12t}$. Calculate $1+\frac{0.08}{12}=1+\frac{2}{300}=\frac{300 + 2}{300}=\frac{302}{300}=\frac{151}{150}$. Then the equation becomes $3.2 = (\frac{151}{150})^{12t}$.

Step3: Take the natural logarithm of both sides

$\ln(3.2)=\ln((\frac{151}{150})^{12t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(3.2)=12t\ln(\frac{151}{150})$.

Step4: Solve for $t$

We know that $\ln(3.2)\approx1.16315$ and $\ln(\frac{151}{150})\approx\ln(1.00667)\approx0.00665$. Then $t=\frac{\ln(3.2)}{12\ln(\frac{151}{150})}$. Substitute the values: $t=\frac{1.16315}{12\times0.00665}=\frac{1.16315}{0.0798}\approx14.6$.

Answer:

14.6