Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
The half life of the sample will be "4 days".
According to the question,
Let,
→ [tex]\frac{N}{N_o} = (\frac{1}{2} )^n[/tex]
By substituting the values, we get
→ [tex]\frac{2}{128} = (\frac{1}{2} )^n[/tex]
→ [tex]\frac{1}{64} = (\frac{1}{2} )^n[/tex]
→ [tex](\frac{1}{2} )^6= (\frac{1}{2} )^n[/tex]
→ [tex]n=6[/tex]
So,
It takes 6 half life just to reduce from 128 mg - 2mg which is also "24 days".
→ [tex]6 \ half \ life = 24 \ days[/tex]
Hence,
→ [tex]Half \ life = \frac{24}{6}[/tex]
[tex]= 4 \ days[/tex]
Thus the answer above is right.
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