Answer:
ln (0.8)/-0.00012 years
Step-by-step explanation:
We can solve the equation as shown below.
\begin{aligned}25\cdot e^{-0.00012t}&=20\\\\ e^{-0.00012t}&=0.8\\\\ -0.00012t&=\ln\left(0.8\right)\\\\ t&=\dfrac{{\,\ln\left(0.8\right)}}{-0.00012}\\\\ \end{aligned}
25⋅e
−0.00012t
e
−0.00012t
−0.00012t
t
=20
=0.8
=ln(0.8)
=
−0.00012
ln(0.8)
It will take \dfrac{{\,\ln\left(0.8\right)}}{-0.00012}
−0.00012
ln(0.8)
start fraction, natural log, left parenthesis, 0, point, 8, right parenthesis, divided by, minus, 0, point, 00012, end fraction years for the amount of Carbon-141414 remaining in the substance to be 20\text{ g}20 g20, start text, space, g, end text.
