m(t)=85e^(-t/409) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/6 of the original ? The response can be written as t=alnb where a and b are integers and a=1
We are solving this :
85(1/6) = 85e^(-t/409) divide through by 85
(1/6) = e^(-t/409) take the natural log of each side
ln (1/60 = ln e^(-t/409) and we can wriite
ln (1/6) = (-t/409)*ln e and ln e = 1, so we can eliminate this
ln (1/6) = -t/ 409 multiply both sides by -409
ln(1/6) * ( -409) = t
(-409)[ ln 1 - ln 6] = t
(-409)(-ln 6) = t
(409)(ln 6) = t = about 732.83 years
