suppose that $6000 is placed in a savings account at an annual rate of 3%, compounded monthly. assuming that…

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 $P = 6000$, $A=7110$, $r = 0.03$ (since $3%=0.03$), and $n = 12$ (compounded monthly). Substitute these values into the formula: $7110=6000(1 +\frac{0.03}{12})^{12t}$.

Step2: Simplify the equation

First, simplify $1+\frac{0.03}{12}=1 + 0.0025=1.0025$. The equation becomes $\frac{7110}{6000}=(1.0025)^{12t}$. $\frac{7110}{6000}=1.185$, so $1.185=(1.0025)^{12t}$.

Step3: Take the natural logarithm of both sides

$\ln(1.185)=\ln((1.0025)^{12t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(1.185)=12t\ln(1.0025)$.

Step4: Solve for $t$

$t=\frac{\ln(1.185)}{12\ln(1.0025)}$. $\ln(1.185)\approx0.1708$ and $\ln(1.0025)\approx0.002497$. $12\ln(1.0025)=12\times0.002497 = 0.029964$. $t=\frac{0.1708}{0.029964}\approx5.70$.

Answer:

$5.70$