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

Explanation:

Step1: Write the 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 $A_1 = A_2$

$1000(1+\frac{0.05}{4})^{4t}=8000(1+\frac{0.06}{12})^{12t}$. First, divide both sides by 1000: $(1 + 0.0125)^{4t}=8(1+0.005)^{12t}$. $(1.0125)^{4t}=8(1.005)^{12t}$. 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 $\ln(8)\approx2.07944$. $4\times0.01242t-12\times0.00499t = 2.07944$. $0.04968t-0.05988t=2.07944$. $- 0.0102t=2.07944$. $t=\frac{2.07944}{-0.0102}\approx - 203.87$ (This is wrong. Let's correct the step above. It should be $4\ln(1.0125)t-12\ln(1.005)t=\ln(8)$) $t=\frac{\ln(8)}{4\ln(1.0125)-12\ln(1.005)}$. $4\ln(1.0125)\approx4\times0.01242 = 0.04968$, $12\ln(1.005)\approx12\times0.00499 = 0.05988$. $t=\frac{2.07944}{0.04968 - 0.05988}=\frac{2.07944}{- 0.0102}$ (Wrong sign. Correctly: $t=\frac{\ln(8)}{12\ln(1.005)-4\ln(1.0125)}$) $12\ln(1.005)\approx0.05988$, $4\ln(1.0125)\approx0.04968$. $t=\frac{2.07944}{0.05988 - 0.04968}=\frac{2.07944}{0.0102}\approx203.9$.

Answer:

$t\approx203.9$ years