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 am

Question

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?

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    2022-01-05T23:14:40+00:00

    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= \frac{1}{3} A_0.

    Simplifying by A_0, we have:

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

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

    t=\log_{0.75}\frac{1}{3}\approx 3.8.

    Answer: 3.8

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