Answer :
The substance's half-life is the time [tex]t[/tex] that satisfies
[tex]\dfrac12=e^{-kt}[/tex]
Solve for [tex]t[/tex]:
[tex]\ln\dfrac12=\ln e^{-kt}[/tex]
[tex]-\ln2=-kt\ln e[/tex]
[tex]\dfrac{\ln2}k=t[/tex]
Which means the half-life is [tex]t\approx4.8[/tex] days.
[tex]\dfrac12=e^{-kt}[/tex]
Solve for [tex]t[/tex]:
[tex]\ln\dfrac12=\ln e^{-kt}[/tex]
[tex]-\ln2=-kt\ln e[/tex]
[tex]\dfrac{\ln2}k=t[/tex]
Which means the half-life is [tex]t\approx4.8[/tex] days.
I'll Be Solving Two Different Questions.
Formula For Both : [tex]N(t)=N_{o}E^{-kt}[/tex]
1. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1455
[tex]\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}\end{array}\right][/tex]
2. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1446.
[tex]\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}Number&1\end{array}\right][/tex]
We Determine That 20 Is Half Of 40.
With That Being Said
[tex]\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}&=20\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}In(E^{-0.1455})=\frac{In\frac{1}{2} }{-0.1455} \end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1455}&Or&T=\frac{-In(2)}{0.1455}\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}-0.5In/0.1455\\=4.76\end{array}\right][/tex]
Rounding It To Be [ 4.8 ]
[tex]\left[\begin{array}{ccc}Number&2\end{array}\right][/tex]
Again 20 Is Determined To Be Half Of 40
With That Being Said
[tex]\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}&=20\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}In(E^{-0.1446})=\frac{In\frac{1}{2} }{-0.1446} \end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1446}&Or&T=\frac{-In(2)}{0.1446}\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}-0.5In/0.1446\\=4.79\end{array}\right][/tex]
Rounding It To Be [ 4.8 ]