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A substance decays so that the amount A of the substance left after t years is given by: A = A 0 · (0.75) t , where A 0 is the original amount of the substance. What is the amount of time it takes for there to be one-third of this substance left, rounded to the nearest tenth of a year?

User Sherard
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Let
A_0 be the original amount of the substance. After t years, the amount left A is given by:


\displaystyle{ A=A_0(0.75)^t.

For example, after 1 year, the amount is


\displaystyle{ A=A_0(0.75)^1=A_0\cdot (0.75)=0.75A_0.

Thus, after one year, only 0.75=3/4 of the original amount is left.


We want to solve for t, such that
\displaystyle{ A_0(0.75)^t is one-third of
A_0, so we set the equation:


\displaystyle{ \displaystyle{ A_0(0.75)^t= (1)/(3) A_0.

Simplifying by
A_0, we have:


\displaystyle{ \displaystyle{(0.75)^t= (1)/(3).

This is an exponential equation, so we can solving it by rewriting it as a logarithm:


t=\log_(0.75)(1)/(3)\approx 3.8.


Answer: 3.8
User Dylan Lacey
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