Question 1156050
<br>
{{{3 = 10(0.5^(t/18.72))}}}
{{{0.3 = 0.5^(t/18.72)}}}
{{{log((0.3)) = (t/18.72)*log((0.5))}}}
{{{t/18.72 = log((0.3))/log((0.5))}}}
{{{t = 18.72*(log((0.3))/log((0.5)))}}} = 32.515995..., or 32.52 to the nearest hundredth.<br>
ANSWER: 32.52 days<br>
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Added comment....<br>
I find that using the given formula {{{A(t) = 10(0.5^(t/18.72))}}} to solve the problem adds an unnecessary level of difficulty to the problem.  I would much rather simply solve for the number of half-lives and multiply that answer by the given length of the half-life.<br>
{{{3 = 10(0.5^x)}}}
{{{0.3 = 0.5^x}}}
{{{log((0.3)) = x*log((0.5))}}}
{{{x = log((0.3))/log((0.5)) = 1.7369656}}}<br>
t = 18.72*1.7369656 = 32.515995....<br>
This method of solving the problem has a huge advantage if you are solving a large number of similar problems involving half-lives.<br>
If the problems are with all different half-lives, then solving the problem by the first method shown involves a different first calculation for every problem; with the second method, the first calculation is always the same, and then you just need to multiply the result of that calculation by the half-life.<br>
Someone who works this kind of problem often would undoubtedly work the problem this way.<br>
For example, suppose we have an initial amount 23 and a final amount 5, with a half-life of 385 years.  The calculations would be<br>
(1) find the number of half-lives:
{{{x = log((5/23))/log((0.5))}}} = 2.201634....
(2) multiply by the half-life:
{{{2.201634*385}}} = 847.63 years<br>