## College Algebra (6th Edition)

In the model for decay, substitute A with $\displaystyle \frac{88}{100}A_{o}$ (88$\%$ of $A_{o})$ and solve for t: $\displaystyle \frac{88}{100}A_{o}=A_{0}e^{-0.000121t}\qquad .../\times\frac{1}{A_{o}}$ $\displaystyle \frac{88}{100}=e^{-0.000121t}\qquad .../$ take ln( ) of both sides $\ln 0.88=-0.000121t\qquad .../\div(-0.000121)$ $t=\displaystyle \frac{\ln 0.88}{-0.000121}\approx$1056.47414471 In 1989, the skeletons were about 1056 years old.