# When answering the following question, explain each step very carefully, specifically any...

## Question:

When answering the following question, explain each step very carefully, specifically any conversions.

Carbon 14 decays exponentially with a half-life of 5750 years, the time it takes for one half of the carbon 14 to decay. If {eq}C_0 {/eq} is the original concentration of carbon 14, which of the following functions correctly approximates C(t), the quantity remaining as a function of time, t, in years?

A. {eq}C(t) = C_0e^{-0.000052t} {/eq}

B. {eq}C(t) = C_0e^{-0.00012t} {/eq}

C. {eq}C(t) = C_0e^{-2875t} {/eq}

D. {eq}C(t) = C_0e^{-5750t} {/eq}

## Evaluating exponential functions

We have a given function with exponential form {eq}y=y_0 e^{At} {/eq}, and we are given some information about it, we can use the information to find the parameters of the function substituting in the equation.

## Answer and Explanation:

The general function has the form:

{eq}$$C(t)=Ae^{Bt}$$ {/eq}

1. We use the initial condition {eq}C(t=0)=C_0 {/eq}, that is:

{eq}\begin{align} C(t)=C_0=Ae^{B\times 0}=A\Rightarrow\;\;A=C_0 \end{align} {/eq}

So far we have that:

{eq}\begin{align} C(t)=C_0e^{Bt} \end{align} {/eq}

2. Now we use the condition that the half-life is the time for a decay of one half. Meaning that when {eq}t=5750 {/eq}, the value of {eq}C(t) {/eq} is {eq}\frac{C_0}{2} {/eq}. That is:

{eq}\begin{align} C(5750)=\frac{C_0}{2}=C_0e^{B\times 5750}\\ e^{5750\,B}=\frac{1}{2}\\ 5750\,B=\ln \frac{1}{2}\\ B=\frac{\ln \frac{1}{2}}{5750}\\ B=-0.00012 \end{align} {/eq}

Therefore the equation is:

{eq}\begin{align} C(t)=C_0e^{-0.00012\,t} \end{align} {/eq}

Then the answer is B