Answer:
12.16 hours
Step-by-step explanation:
I'm going to assume my assumption is right.
You can let me know differently.
[tex]A=A_0 \cdot e^{-0.057t}[/tex]
We want to know the time, t, such that A will be half it's initial population.
This means we want to solve the following equation for t:
[tex]\frac{A_0}{2}=A_0 \cdot e^{-0.057t}[/tex]
Divide both sides by [tex]A_0[/tex]:
[tex]\frac{1}{2}=e^{-0.057t}[/tex]
Now we are ready to try to get the variable by itself.
Let's rewrite in the equivalent logarithm form:
[tex]\ln(\frac{1}{2})=-0.057t[/tex]
Divide both sides by -0.057:
[tex]\frac{\ln(\frac{1}{2})}{-0.057}=t[/tex]
Put left hand side into a calculator:
[tex]12.16=t[/tex]
So the half-life is 12.16.
This means that the initial population will be brought down to half of the initial population in 12.16 hours approximately.
