## College Algebra (6th Edition)

about $15,679$ years old
In the model for decay, substitute A with $\displaystyle \frac{15}{100}A_{o}$ (15$\%$ of $A_{o})$ and solve for t: $\displaystyle \frac{15}{100}A_{o} =A_{0}e^{-0.000121t}\displaystyle \qquad .../\times\frac{1}{A_{o}}$ $\displaystyle \frac{15}{100}=e^{-0.000121t}\qquad .../$ take ln( ) of both sides $\ln 0.15=-0.000121t\qquad .../\div(-0.000121)$ $t=\displaystyle \frac{\ln 0.15}{-0.000121}\approx$15,678.677561 The paintings are about $15,679$ years old.