1. ## Help with Algebra

A 100 mg sample of radioactive actinium 227 decays according this equation:

$\displaystyle N=100e^{-0.03194 t}$

When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

Is this right?:

$\displaystyle 50=100e^{-0.03194 t}$

$\displaystyle 0.5=e^{-0.03194 t}$

I know I have to use logs, but I don't know how to finish because of the $\displaystyle e$ .

2. Originally Posted by Dex

A 100 mg sample of radioactive actinium 227 decays according this equation:

$\displaystyle N=100e^{-0.03194 t}$

When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

Is this right?:

$\displaystyle 50=100e^{-0.03194 t}$

$\displaystyle 0.5=e^{-0.03194 t}$

I know I have to use logs, but I don't know how to finish because of the $\displaystyle e$ .

$\displaystyle 0.5=e^{-0.03194 t}$

Take natural logs:

$\displaystyle \ln(0.5)=-0.03194 t$

CB

3. Originally Posted by Dex

A 100 mg sample of radioactive actinium 227 decays according this equation:

$\displaystyle N=100e^{-0.03194 t}$

When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

Is this right?:

$\displaystyle 50=100e^{-0.03194 t}$

$\displaystyle 0.5=e^{-0.03194 t}$

I know I have to use logs, but I don't know how to finish because of the $\displaystyle e$ .

$\displaystyle 0.5=e^{-0.03194 t}~\implies~\ln(0.5)=\ln\left(e^{-0.03194 t}\right)~\implies~\underbrace{-\ln(2)}_{by\ calculator}=-0.03194 t$