Answer
The parchment is about 2488 years old.
Work Step by Step
We can find the value of $k$:
$m(t) = m(0)e^{kt}$
$m(5730) = m(0)e^{5730k} = 0.5~m(0)$
$e^{5730k} = 0.5$
$5730~k = ln(0.5)$
$k = \frac{ln(0.5)}{5730}$
$k = -0.000121$
Then:
$m(t) = m(0)~e^{-0.000121~t}$
We can find the time $t$ when only 74% remains:
$m(t) = m(0)~e^{-0.000121~t} = 0.74~m(0)$
$e^{-0.000121~t} = 0.74$
$(-0.000121)~t = ln(0.74)$
$t = \frac{ln(0.74)}{-0.000121}$
$t = 2488~years$
The parchment is about 2488 years old.
