#### Answer

about $15,679$ years old

#### Work Step by Step

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.