Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Exercises - Page 426: 93

$t=1.833~years\approx669~days$

$A=P(1+\frac{r}{n})^{nt}$ $r=5.2$% $=0.052$ $P=100,00$ $A=100,000+10,000=110,000$ $110,000=100,000(1+\frac{0.052}{365})^{365t}$ $1.1=(1+\frac{0.052}{365})^{365t}$ $\log1.1=\log(1+\frac{0.052}{365})^{365t}$ $\log1.1=365t[\log(1+\frac{0.052}{365})]$ $t=\frac{\log1.1}{365\log(1+\frac{0.052}{365})}=1.833$