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

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


`\frac12=e^(-kt)`

Solve for
`t`:


`\ln\frac12=\ln e^(-kt)`

`-\ln2=-kt\ln e`

`\frac{\ln2}k=t`

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

Answer 2

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))=(In(1)/(2) )/(-0.1455) \end{array}\right]`

`\left[\begin{array}{ccc}T=(In(1)/(2) )/(-0.1455)&Or&T=(-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))=(In(1)/(2) )/(-0.1446) \end{array}\right]`

`\left[\begin{array}{ccc}T=(In(1)/(2) )/(-0.1446)&Or&T=(-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 ]