two bank accounts are opened at the same time. the first has a principal of $1000 in an account earning 6%…
two bank accounts are opened at the same time. the first has a principal of $1000 in an account earning 6% compounded quarterly. the second has a principal of $4000 in an account earning 4% interest compounded monthly. determine the number of years, to the nearest tenth, at which the account balances will be equal. use the compound - interest formula, $a = pcdot(1+\frac{r}{n})^{nt}$ to find the number of years at which the account balances will be equal.
Answer
Explanation:
Step1: Set up the compound - interest equations
For the first account: $P_1 = 1000$, $r_1=0.06$, $n_1 = 4$. So, $A_1=1000(1 +\frac{0.06}{4})^{4t}$. For the second account: $P_2 = 4000$, $r_2=0.04$, $n_2 = 12$. So, $A_2=4000(1+\frac{0.04}{12})^{12t}$. Set $A_1 = A_2$: $1000(1 +\frac{0.06}{4})^{4t}=4000(1+\frac{0.04}{12})^{12t}$.
Step2: Simplify the equation
Divide both sides by 1000: $(1 + 0.015)^{4t}=4(1+\frac{0.04}{12})^{12t}$. $(1.015)^{4t}=4(1+\frac{1}{300})^{12t}$. $(1.015)^{4t}=4( \frac{301}{300})^{12t}$. Take the natural logarithm of both sides: $4t\ln(1.015)=\ln(4)+12t\ln(\frac{301}{300})$.
Step3: Solve for t
$4t\ln(1.015)-12t\ln(\frac{301}{300})=\ln(4)$. Factor out t: $t(4\ln(1.015)-12\ln(\frac{301}{300}))=\ln(4)$. $4\ln(1.015)\approx4\times0.014889 = 0.059556$. $12\ln(\frac{301}{300})\approx12\times0.003322 = 0.039864$. $4\ln(1.015)-12\ln(\frac{301}{300})=0.059556 - 0.039864=0.019692$. $t=\frac{\ln(4)}{4\ln(1.015)-12\ln(\frac{301}{300})}$. Since $\ln(4)\approx1.3863$, then $t=\frac{1.3863}{0.019692}\approx70.4$.
Answer:
$70.4$