13.5k views
1 vote
Logarithm 4) 20 grams of an unknown compound decays continuously according to the model A = A0e^-kt where A is the amount of compound remaining after + years. If the compound decays at a yearlyrate of 2%, how long before the amount of compound reaches one-fourth of its originalamount?

Logarithm 4) 20 grams of an unknown compound decays continuously according to the-example-1

1 Answer

3 votes

We know that the function describing the amount remaining is given by:


A=A_0e^(-kt)

In this case the initial amount is 20 and the decay rate is 0.02 in decimal form, hence the function we need is:


A=20e^(-0.02t)

We want to know how long until the compound reaches one fourth of the original amount, this means that we need to find the time it takes to have 5 gr of the compond. Plugging this value in the function and solving for t we have:


\begin{gathered} 5=20e^(-0.02t) \\ e^(-0.02t)=(5)/(20) \\ e^(-0.02t)=(1)/(4) \\ \ln e^(-0.02t)=\ln ((1)/(4)) \\ -0.02t=\ln ((1)/(4)) \\ t=-(1)/(0.02)\ln ((1)/(4)) \\ t=69.31 \end{gathered}

Therefore it takes approximately 69 years to have one forth of the orginal amount.

User Teemu Tapanila
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories