assume $11,000 grows to $16,000 in an investment that earns interest at the rate of 5%/year compounded…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{m})^{mt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $m$ is the number of times interest is compounded per year, and $t$ is the number of years. We are given $A = 16000$, $P = 11000$, $r=0.05$, and $m = 12$. Substitute these values into the formula: $16000=11000(1 +\frac{0.05}{12})^{12t}$.

Step2: Simplify the equation

First, divide both sides of the equation by $11000$: $\frac{16000}{11000}=(1+\frac{0.05}{12})^{12t}$, which simplifies to $\frac{16}{11}=(1+\frac{0.05}{12})^{12t}$. Calculate $1+\frac{0.05}{12}=1+\frac{1}{240}=\frac{240 + 1}{240}=\frac{241}{240}$. So, $\frac{16}{11}=(\frac{241}{240})^{12t}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{16}{11})=\ln((\frac{241}{240})^{12t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{16}{11}) = 12t\ln(\frac{241}{240})$. We know that $\ln(\frac{16}{11})\approx\ln(16)-\ln(11)\approx2.7726 - 2.3979=0.3747$ and $\ln(\frac{241}{240})\approx\ln(241)-\ln(240)\approx5.4849 - 5.4806 = 0.0043$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{16}{11})}{12\ln(\frac{241}{240})}$. Substitute the values of the logarithms: $t=\frac{0.3747}{12\times0.0043}=\frac{0.3747}{0.0516}\approx7.26$.

Answer:

$7.26$