Question

A substance decays according to A = Ae-0.057, where I is in hours and Ao is the initial amount. Determine the half-life of the substance. Round your answer to two
decimal places, if necessary.

Answer 1

Answer:

12.16 hours

Step-by-step explanation:

I'm going to assume my assumption is right.

You can let me know differently.

$A=A_0 \cdot e^{-0.057t}$

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:

$\frac{A_0}{2}=A_0 \cdot e^{-0.057t}$

Divide both sides by $A_0$:

$\frac{1}{2}=e^{-0.057t}$

Now we are ready to try to get the variable by itself.

Let's rewrite in the equivalent logarithm form:

$\ln(\frac{1}{2})=-0.057t$

Divide both sides by -0.057:

$\frac{\ln(\frac{1}{2})}{-0.057}=t$

Put left hand side into a calculator:

$12.16=t$

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.