Chapter 8 - Section 8.6 - Systems of Nonlinear Equations - 8.6 Assess Your Understanding - Page 617: 105

$r=0.06$

$A(t)=P(1+\frac{r}{n})^{nt},$ $A(3)=1800, t=3, P=1500, n=4$ Thus, $1800=1500(1+\frac{r}{4})^{12},$ $\ln 1.2=12\ln {(1+\frac{r}{4})},$ $0.0152=\ln {(1+\frac{r}{4})},$ $e^{0.0152}=1+\frac{r}{4},$ $0.015\approx \frac{r}{4},$ $r\approx 0.06$