Question 173403
the formula is:
{{{p[t] = p[0]*e^(-k*t)}}}
that minus k is important since the exponent has to be negative in a decay problem.
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you are given:
decay rate is 6.3% per day.
this translates to .063 per day which is equal to k since k, in this case, is the decay rate per day.
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you are also asked for the half life.
this means at what point in time will the amount of krypton be half of what it is today.
if {{{p[0]}}} = 1 = the amount of krypton today, then {{{p[t]}}} = .5 is the amount of krypton when one half life has occurred.
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your equation:
{{{p[t] = p[0]*e^(-k*t)}}}
becomes:
{{{.5 = 1 * e^(-.063*t)}}}
which becomes:
{{{.5 = e^(-.063*t)}}}
which you can solve using the log or ln function of your calculator.
we'll use the log function.
either one will get the same answer.
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your equation becomes:
{{{log(10,.5) = log(10,e^(-.063*t))}}}
since {{{log(10,b^a) = a*log(10,b)}}} your equation becomes:
{{{log(10,.5) = -.063*t*log(10,e)}}}
if you divide both sides of this equation by {{{-.063*log(10,e)}}}, you get:
{{{log(10,.5)/(-.063*log(10,e)) = t}}}
which you can solve using your calculator to get:
t = 11.0023362 days
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to prove this is true, plug that value in your original equation for t.
that equation:
{{{p[t] = p[0]*e^(-k*t)}}}
becomes:
{{{.5 = 1*e^(-.063*11.0023362)}}}
which becomes:
{{{.5 = e^(-.693145181)}}}
which becomes:
.5 = .5
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value is good.
answer is:
t = 11.0023362