Alright, let's solve this problem step by step.
1. Identify key information:
- Initial amount of the isotope: [tex]\(120 \, \text{mg}\)[/tex]
- Remaining threshold amount: [tex]\(30 \, \text{mg}\)[/tex]
- Daily decay rate: [tex]\(16\% \)[/tex]
2. Convert the decay rate into a decay factor:
- Decay rate of [tex]\(16\% \)[/tex] means the isotope retains [tex]\(100\% - 16\% = 84\% \)[/tex] of its mass each day.
- Therefore, the decay factor is [tex]\(0.84\)[/tex].
3. Formulate an inequality:
- The problem asks for the number of days [tex]\(t\)[/tex] such that the remaining amount of the substance is less than [tex]\(30 \, \text{mg}\)[/tex].
- Hence, the inequality will be:
[tex]\[
120 \cdot (0.84)^t < 30
\][/tex]
4. Solve the inequality:
- We need to isolate [tex]\(t\)[/tex].
- Divide both sides of the inequality by 120:
[tex]\[
(0.84)^t < \frac{30}{120}
\][/tex]
[tex]\[
(0.84)^t < 0.25
\][/tex]
- Take the logarithm of both sides to solve for [tex]\(t\)[/tex]:
[tex]\[
\log((0.84)^t) < \log(0.25)
\][/tex]
[tex]\[
t \cdot \log(0.84) < \log(0.25)
\][/tex]
- Since [tex]\(\log(0.84)\)[/tex] is negative, divide both sides by [tex]\(\log(0.84)\)[/tex] (remembering to reverse the inequality):
[tex]\[
t > \frac{\log(0.25)}{\log(0.84)}
\][/tex]
5. Calculate the value of [tex]\(t\)[/tex]:
- Using logarithm values:
[tex]\[
t > \frac{\log(0.25)}{\log(0.84)}
\approx 7.95
\][/tex]
6. Interpret the result:
- The smallest integer greater than [tex]\(7.95 \)[/tex] is [tex]\(8 \)[/tex].
So, the correct inequality is:
[tex]\[
120(0.84)^t < 30
\][/tex]
and [tex]\(t\)[/tex] will be around [tex]\(8\)[/tex] days for the sample size remaining to be less than [tex]\(30 \, \text{mg}\)[/tex].
Therefore, the correct answer is:
[tex]\[
120(0.84)^t < 30 ; t = 8 \text{ days}
\][/tex]
