SOLUTION: A certain isotope decays at a rate of 2%per 100 years. If t represent the time in years and y represent the amount of the isotope left then the equation for the situation is y=y0e^

Question 402058: A certain isotope decays at a rate of 2%per 100 years. If t represent the time in years and y represent the amount of the isotope left then the equation for the situation is y=y0e^-0.0002t . In how many years will there be 86% of the isotope left?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A certain isotope decays at a rate of 2%per 100 years. If t represent the time in years and y represent the amount of the isotope left then the equation for the situation is y=y0e^-0.0002t . In how many years will there be 86% of the isotope left?
.
Let x = initial amount
then our formula is
.86x = xe^(-0.0002t)
.
Notice, if we divide both sides by x, we eliminate our unknown:
.86 = e^(-0.0002t)
Solving for x, we take the ln of both sides:
ln(.86) = -0.0002t
ln(.86)/(-0.0002) = t
754.114 years = t

RELATED QUESTIONS
Solve the problem. Use the formula N = Iekt, where N is the number of items in terms... (answered by ankor@dixie-net.com)

In the formula A(t) = A0ekt, A is the amount of radioactive material remaining from an... (answered by scott8148)

Use the formula N = Ie^(kt), where N is the number of items in terms of the initial... (answered by Theo)

Use the formula N = Iekt, where N is the number of items in terms of the initial... (answered by stanbon)

Use the formula N = Iekt, where N is the number of items in terms of the initial... (answered by nerdybill)

A certain isotope of silver, 110Ag decays exponentially at a rate of 2.8% per second.... (answered by Alan3354)

a certain radio active isotope decays at a rate of .2% annually. Determine the half life... (answered by scott8148)

Could someone please help me with this problem? Use the formula N = Ie^kt, where N is the (answered by stanbon)

A certain radioactive isotope decays at a rate of 1% per year; 3.4 kilograms of this... (answered by stanbon)