Chapter 7 Exponential and Logarithmic Functions - Mixed Review of Problem Solving - Lessons 7.1-7.4 - Page 506: 4

3 years

Given $A=Pe^{rt}$ and $r=0.04\\P=2000$ Substitute $A=2000e^{0.04\times 2}\\=2000e^{0.08}\\\approx166.6$ Let $A\leq 2250\\2000e^{0.04x}\leq2250\\e^{0.04x}\leq1.125\\\ln e^{0.04x}\leq\ln1.125\\x\leq2.945$ Thus, we have to wait 3 years.