Question 200908
Remember,  x = x(0) * 2^(-t/h),,,,where x(0) is  initial  amount  at t=0,,t=time(yrs),   and h = half  life (yrs)
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x = x(0) *2^(-t/5.3),,,,,,,(a)
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subst  in given  values
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10 = 15 *2^( - t/5.3)
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10/15 =2/3 = 2^(-t/5.3)
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take  log  both  sides,,,,,remember log (a^b)  = b(loga)
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log (2/3) = log (2^(-t/5.3) ) = (-t/5.3)log2
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log(2/3) / log2  = (-t/5.3)
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-.585 = - t/5.3
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3.10 = t,,,,,,,,,,(b)
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check
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10 = 15 *2^(-3.1/5.3) =10,,,,,,ok
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