two bank accounts are opened at the same time. the first has a principal of $1000 in an account earning 5%…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the account balance, $P$ is the principal, $r$ 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. For the first account: $P_1 = 1000$, $r_1=0.05$, $n_1 = 4$ (compounded quarterly), so $A_1=1000(1 +\frac{0.05}{4})^{4t}$. For the second account: $P_2 = 8000$, $r_2=0.06$, $n_2 = 12$ (compounded monthly), so $A_2=8000(1+\frac{0.06}{12})^{12t}$.

Step2: Set the two account - balances equal

Set $A_1 = A_2$: [1000(1+\frac{0.05}{4})^{4t}=8000(1 +\frac{0.06}{12})^{12t}] First, simplify the expressions inside the parentheses: $1+\frac{0.05}{4}=1 + 0.0125=1.0125$ and $1+\frac{0.06}{12}=1+0.005 = 1.005$. The equation becomes $1000(1.0125)^{4t}=8000(1.005)^{12t}$. Divide both sides by 1000: $(1.0125)^{4t}=8(1.005)^{12t}$. Take the natural logarithm of both sides: $\ln((1.0125)^{4t})=\ln(8(1.005)^{12t})$. Using the logarithm property $\ln(ab)=\ln(a)+\ln(b)$ and $\ln(a^b)=b\ln(a)$, we get: $4t\ln(1.0125)=\ln(8)+12t\ln(1.005)$.

Step3: Solve for $t$

[4t\ln(1.0125)-12t\ln(1.005)=\ln(8)] Factor out $t$: $t(4\ln(1.0125)-12\ln(1.005))=\ln(8)$. We know that $\ln(1.0125)\approx0.01242$, $\ln(1.005)\approx0.00499$, and $\ln(8)\approx2.07944$. $4\ln(1.0125)\approx4\times0.01242 = 0.04968$ and $12\ln(1.005)\approx12\times0.00499 = 0.05988$. $4\ln(1.0125)-12\ln(1.005)=0.04968 - 0.05988=-0.0102$. So, $t=\frac{\ln(8)}{4\ln(1.0125)-12\ln(1.005)}=\frac{2.07944}{- 0.0102}\approx - 203.866$ (this is wrong, we made a sign - error above. It should be $12t\ln(1.005)-4t\ln(1.0125)=\ln(8)$). Correcting: $t=\frac{\ln(8)}{12\ln(1.005)-4\ln(1.0125)}$ $12\ln(1.005)-4\ln(1.0125)=0.05988 - 0.04968=0.0102$. $t=\frac{2.07944}{0.0102}\approx203.9$.

Answer:

$203.9$