Help with Algebra

**Dex** · September 19th 2010, 10:12 PM

Please Help me answer this question, I received my exam back but got this question wrong.

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

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

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?:

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

$0.5=e^{-0.03194 t}$

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

Please help!

**CaptainBlack** · September 19th 2010, 10:45 PM

Originally Posted by Dex

Please Help me answer this question, I received my exam back but got this question wrong.

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

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

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?:

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

$0.5=e^{-0.03194 t}$

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

Please help!

$0.5=e^{-0.03194 t}$

Take natural logs:

$\ln(0.5)=-0.03194 t$

CB

**earboth** · September 19th 2010, 10:46 PM

Originally Posted by Dex

Please Help me answer this question, I received my exam back but got this question wrong.

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

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

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?:

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

$0.5=e^{-0.03194 t}$

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

Please help!

1. Logarithms to the bas e e are called natural logarithms. Your calculator is able to calculate the corresponding values. Use the function ln().

2. Taking the natural logs on both sides will give:

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

3. Solve for t.

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