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?

Answer 1

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