Question

a 40 gram sample of a substance thats used for drug research has a k-value of 0.1455 . find the substance half-life in days round your answer to the nearest tenth

Answer 1

I'll Be Solving Two Different Questions.

Formula For Both : $N(t)=N_{o}E^{-kt}$

1. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1455

$\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}\end{array}\right]$

2. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1446.

$\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}\end{array}\right]$

$\left[\begin{array}{ccc}Number&1\end{array}\right]$

We Determine That 20 Is Half Of 40.

With That Being Said

$\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}&=20\end{array}\right]$

$\left[\begin{array}{ccc}In(E^{-0.1455})=\frac{In\frac{1}{2} }{-0.1455} \end{array}\right]$

$\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1455}&Or&T=\frac{-In(2)}{0.1455}\end{array}\right]$

$\left[\begin{array}{ccc}-0.5In/0.1455\\=4.76\end{array}\right]$

Rounding It To Be [ 4.8 ]

$\left[\begin{array}{ccc}Number&2\end{array}\right]$

Again 20 Is Determined To Be Half Of 40

With That Being Said

$\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}&=20\end{array}\right]$

$\left[\begin{array}{ccc}In(E^{-0.1446})=\frac{In\frac{1}{2} }{-0.1446} \end{array}\right]$

$\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1446}&Or&T=\frac{-In(2)}{0.1446}\end{array}\right]$

$\left[\begin{array}{ccc}-0.5In/0.1446\\=4.79\end{array}\right]$

Rounding It To Be [ 4.8 ]

Answer 2

The substance's half-life is the time $t$ that satisfies

$\dfrac12=e^{-kt}$

Solve for $t$:

$\ln\dfrac12=\ln e^{-kt}$
$-\ln2=-kt\ln e$
$\dfrac{\ln2}k=t$

Which means the half-life is $t\approx4.8$ days.