Question

Practice Exam 1, Question 10. Why does e^-rt get reduced to 1? i.e. removed from the answer?

Answer 1

\[A=A _{o} e^{-rt} \]
formulate an equation in terms of t, time to find out how long it will take for A to fall to \[\frac{ 1 }{ 4 } A _{o}\], the initial amount.

we state equation for the quarter life by setting \[ A=(1/4)A_{o}\] and substituting \[ (1/4)A_{o}\] in for A.
this eliminates the variable a and the question defines r as a constant.
so \[\frac{ A _{o} }{ 4 }=A_{o}e ^{-rt} \to 1/4=e ^{-rt} \to \ln \frac{ 1 }{ 4 } =\ln e ^{-rt}\]
we just eliminated the variable Ao by dividing both sides by Ao,
and we then eliminated the exponent for of e by takingthe natural log of both sides.\[\ln (e ^{x})=x \because \ln e=1 \because e ^{1}=e \]
to simplify the form:
\[\ln(1)-\ln(4)=-rt\]
by the properties of logs. we know that ln(1)=0 by definition, so
\[- \ln(4) = -rt\]
and we divide both sides by -r to solve for t:
\[\frac{ \ln 4 }{ r }\] which is our answer to part a.

to solve for part b we differentiate the function as so:
\[dA/dt= A _{o}e ^{-rt}*-r\]
because Ao and r are constandts and e^x is the derivitve of itself. then by chain rule you multiply that by the derivitive of -rt which is -r.
and then set \[t=\frac{ \ln 4 }{ r }\]

which gives us:
\[\frac{ dA }{ dt }=-rA _{o}e ^{-r \frac{ \ln4 }{ r }}\]
the r's cancel out in the exponent leaving \[\frac{ dA }{ dt }=-rA _{o}e^{-ln4}\]
\[e ^{-\ln4} \to \frac{ 1 }{ e ^{\ln4} } \to \frac{ 1 }{ 4 }\]
again because e ^lnx = x ;

which leaves \[\frac{ -rA _{o} }{ 4 }\]
as our final answer to the second part.

just remmember ln e = 1; e^lnx=x; and ln(ab)=lna+lnb, and also ln a^b=blna.

hope this helps.

Answer 2

Ah ha. Excellent. Thank you. The answer in the first answer key appears to be wrong or at least poorly written. I hadn't bothered to cross reference the two PDFs. Thanks for taking the time to answer that.