<PRE><B>Question 1183394</B>
p(t) = 20 * e^(-.00012t)
p(0) = 20
p(t) = .6 * 20 = 12


p(t) = e^(-.00012t) = 12
formula becomes:
12 = 20 * e^(-.00012t)
divide both sides of the equation by 20 to get:
12/20 = e^(-.00012 * t)
take the natural log of both sides of the equation to get:
ln(12/20) = ln(e^(-.00012 * t)
since, in general, ln(e^x) = x * ln(e), the formula becomes:
ln(12/20) = -.00012 * t * ln(e)
since ln(e) = 1, the equation becomes:
ln(12/20) = -.00012 * t
solve for t to get:
t = ln(12/20) / -.00012 = 4256.880198 years.


confirm by replacing t in the original equation to get:
p(4256.880198) = 20 * e^(-.00012 * 4256.880198) = 12.
this confirms the value of t is correct.



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