question 10\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 form), $n$ 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). Substitute these values into the formula: $320000=100000(1 +\frac{0.08}{12})^{12t}$.

Step2: Simplify the equation

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

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 $t=\frac{\ln(3.2)}{12\ln(\frac{151}{150})}$ $12\ln(\frac{151}{150})\approx12\times0.00665 = 0.0798$ $t=\frac{1.16315}{0.0798}\approx14.6$

Answer:

$14.6$