Answer
(a) $42.8\ yrs$
(b) $42.7\ yrs$
Work Step by Step
Given $P=5000, A=18000, r=0.03$, we have:
(a) $n=12$, use $A=P(1+\frac{r}{n})^{nt}$, we have $18000=5000(1+\frac{0.03}{12})^{12t} \longrightarrow (1.0025)^{12t}=3.6 \longrightarrow 12t\ ln(1.0025)=ln(3.6)$, thus $t=\frac{ln(3.6)}{12ln(1.0025)}\approx42.8\ yrs$
(b) Use $A=Pe^{rt}$, we have $18000=5000e^{0.03t} \longrightarrow e^{0.03t}=3.6 \longrightarrow 0.03t=ln(3.6)$, thus $t=\frac{ln(3.6)}{0.03}\approx42.7\ yrs$
