Answer
$t = 19.7 d$
Work Step by Step
Let $N_{AA0}$ be the number of element $AA$ at $t = 0$. At a later time t due to decay, we have
$N_{AA0} = N_{AA} + N_{BB} +N_{CC}$
Where
$\frac{N_{BB}}{N_{CC}} = 2.00 $
$\frac{N_{CC}}{N_{AA}} = 1.50$
$\frac{N_{BB}}{N_{AA}} = 3.00$
Substitute the values into the equation. After rearranged, the equation becomes
$N_{AA0} = N_{AA} + 3.00N_{AA} + 1.50N_{AA}$
$N_{AA0} = 5.50 N_{AA}$
$\frac{N_{AA0}}{N_{AA}} = 5.50 $
The equation of radioactive dating,
$N_{AA} = N_{AA0} e^{-\lambda t}$ and $\lambda = \frac{ln2}{T_{1/2}}$
$N_{AA} = N_{AA0} e^{-(ln2/T_{1/2}) t}$
Solve for t and rearrange the equation,
$t = -ln(\frac{N_{AA}}{N_{AA0}}) (\frac{T_{1/2}}{ln2})$
$t = ln(\frac{N_{AA0}}{N_{AA}}) (\frac{T_{1/2}}{ln2})$
Substitute all values into equation
$t = ln(5.50) (\frac{8 d}{0.693})$
$t = 19.7 d$
