Question

See question below. Round your answer to the nearest tenth.

Answer 1

`8.0\:hours`

1) In this problem, we must remind that the final amount on the left is half the initial one. Thus, we can plug into that formula the initial amount as 1 and the final 0.5

2) So, let's plug them into the formula:

`\begin{gathered} y=y_0e^(-0.086t) \\ \\ 0.5=1\cdot e^(-0.086t) \\ \\ e^(-0.086t)=0.5 \\ \\ \ln \left(e^(-0.086t)\right)=\ln \left(0.5\right) \\ \\ -0.086t=\ln \left(0.5\right) \\ \\ (-0.086t)/(-0.086)=(\ln \left(0.5\right))/(-0.086) \\ \\ t=-(\ln \left(0.5\right))/(0.086) \\ \\ t\approx8.0 \end{gathered}`

Thus, the half-life is 8 hours