Rate of decay is proportional to amount of substance remaining. M=Ce^(kt). Time is in days. Initially 125.3 grams of substance.after 3 days,98.1 grams remain. What is constant k?

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Answer 1 · 2018-03-02T15:25:42

#k=1/3*ln(98.1/123.5)#

First, we need to determine #C#. This can be done by solving the equation at time #t=0#, since we are given the initial amount of the substance.

Solve for #C#:

#C=M*e^(-kt)#

With #M=123.5g#, we have at #t=0#:

#C=123.5*e^0#

#=>C=123.5g#.

For a later time, we are given #t=3# days, #M=98.1"g",# and we now know #C#. First, we can solve for #k#.

Begin by taking the natural log (ln) of both sides, the inverse of #e#:

#ln(M)=ln(Ce^(kt))#

By the product rule for logarithms, this is equivalent to:

#ln(M) = ln(C) + ln(e^(kt))#

Because #e# and #ln# are inverses and "undo" each other:

#=>ln(M) = ln(C) + (kt)#

#=>kt = ln(M)-ln(C)#

Applying the quotient rule for logs:

#=>kt=ln(M/C)#

Then:

#=>k=t^-1ln(M/C)#

Using our known values:

#k=1/3*ln(98.1/123.5)#

Note that this gives #k# in units of inverse days, so values of #t# must have units of days from now on.