Given:
Initial value = 9 grams
k-value = 0.1459
Let's find the substance's half-life in days.
Apply the formula:
[tex]N=N_0e^{-kt}[/tex]
The half-life of a substance can be expressed as:
[tex]\begin{gathered} \frac{N_0}{2}=N_0e^{-kt} \\ \\ \frac{1}{2}=e^{-kt} \end{gathered}[/tex]
Plug in 0.1459 for k and solve for t.
We have:
[tex]\begin{gathered} \frac{1}{2}=e^{-0.1459t} \\ \\ \end{gathered}[/tex]
Take the natural logarithm of both sides:
[tex]\begin{gathered} ln(\frac{1}{2})=-0.1459t\text{ ln\lparen e\rparen} \\ \\ −0.693147=-0.1459t \\ \\ t=\frac{−0.693147}{-0.1459} \\ \\ t=4.75\approx4.8\text{ days} \end{gathered}[/tex]
Therefore, the substance's half-life in days is 4.8 days.
ANSWER:
4.8 days
