Here are the variables:
[tex]n_{0}=29\text{ grams}\\
n=14.5\text{ grams}\\
k\approx.1342[/tex]
Here are the details:
[tex]14.5=29e^{-.1342t}\\ .5=e^{-.1342t}\\ \ln(.5)=\ln(e^{-.1342})\\ \ln(.5)=-.1342t\\ ^{\ln(.5)}/_{-.1342}=t\\ 5.2\approx t[/tex]
Here is your answer:
[tex]5.2\approx t[/tex]
