two bank accounts are opened at the same time. the first has a principal of $1000 in an account earning 12%…

Explanation:

Step1: Write 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: $P_1 = 1000$, $r_1=0.12$, $n_1 = 4$, so $A_1=1000(1 +\frac{0.12}{4})^{4t}=1000(1 + 0.03)^{4t}=1000(1.03)^{4t}$. For the second account: $P_2 = 3000$, $r_2=0.06$, $n_2 = 12$, so $A_2=3000(1+\frac{0.06}{12})^{12t}=3000(1 + 0.005)^{12t}=3000(1.005)^{12t}$.

Step2: Set $A_1 = A_2$

$1000(1.03)^{4t}=3000(1.005)^{12t}$. Divide both sides by 1000: $(1.03)^{4t}=3(1.005)^{12t}$. Take the natural logarithm of both sides: $\ln((1.03)^{4t})=\ln(3(1.005)^{12t})$. Using the property of logarithms $\ln(ab)=\ln(a)+\ln(b)$ and $\ln(a^b)=b\ln(a)$, we get $4t\ln(1.03)=\ln(3)+12t\ln(1.005)$.

Step3: Solve for $t$

$4t\ln(1.03)-12t\ln(1.005)=\ln(3)$. Factor out $t$: $t(4\ln(1.03)-12\ln(1.005))=\ln(3)$. $4\ln(1.03)\approx4\times0.0296=0.1184$, $12\ln(1.005)\approx12\times0.00499 = 0.0599$. $4\ln(1.03)-12\ln(1.005)=0.1184 - 0.0599=0.0585$. $t=\frac{\ln(3)}{4\ln(1.03)-12\ln(1.005)}=\frac{1.0986}{0.0585}\approx18.8$.

Answer:

$18.8$