Question

a certain isotope decays so that the amount A remaining after t years is given by: A=A0 x e^-0.03t, where A0 is the original amount of the isotope. to the nearest year, the half-life of the isotope (the amount of the time it takes to decay to half the original amount) is ____ years

Answer 1

Answer:

$t = 23.10\ years$

Step-by-step explanation:

We have an exponential decay function:

$A = A_0e ^{-0.03t}$

This function decreases as the years pass. That is, if A is the quantity for a time t and $A_0$ is the original quantity at time t = 0 then:

$A$

We want to know for what value of t the value of A decreases by half of  $A_0$. That is, we want to know when $A = \frac{A_0}{2}$

Then we do:

$\frac{A_0}{2} = A_0e ^{-0.03t}\\\\\frac{1}{2} =e ^{-0.03t}\\\\ln(\frac{1}{2}) = -0.03t\\\\t = \frac{ln(\frac{1}{2})}{-0.03}$

$t = 23.10\ years$