Answer:
The substance's half-life is 6.4 days
Step-by-step explanation:
Recall that the half life of a substance is given by the time it takes for the substance to reduce to half of its initial amount. So in this case, where they give you the constant k (0.1088) in the exponential form:
[tex]N=N_0\,e^{-k\,*\,t}[/tex]
we can replace k by its value, and solve for the time "t" needed for the initial amount [tex]N_0[/tex] to reduce to half of its value ([tex]N_0/2[/tex]). Since the unknown resides in the exponent, to solve the equation we need to apply the natural logarithm:
[tex]N=N_0\,e^{-k\,*\,t}\\\frac{N_0}{2} =N_0\,e^{-0.1088\,*\,t}\\\frac{N_0}{2\,*N_0} =e^{-0.1088\,*\,t}\\\frac{1}{2} =e^{-0.1088\,*\,t}\\ln(\frac{1}{2} )=-0.1088\,t\\t=\frac{ln(\frac{1}{2} )}{-0.1088} \\t=6.37\,\,days[/tex]
which rounded to the nearest tenth is: 6.4 days
