The half-life of the element is 13 minutes, which means after 13 minutes, any starting amount decays to half. So element X decays with a rate k such that
[tex]\dfrac12 = e^{13k}[/tex]
Solve for k :
[tex]\ln\left(\dfrac12\right) = \ln\left(e^{13k}\right)[/tex]
[tex]-\ln(2) = 13k \ln(e)[/tex]
[tex]-\ln(2) = 13k[/tex]
[tex]\implies k = -\dfrac{\ln(2)}{13}[/tex]
Now, we solve for t such that
[tex]36 = 710e^{kt}[/tex]
[tex]e^{kt} = \dfrac{18}{355}[/tex]
[tex]\ln\left(e^{kt}\right) = \ln\left(\dfrac{18}{355}\right)[/tex]
[tex]kt = \ln\left(\dfrac{18}{355}\right)[/tex]
[tex]-\dfrac{\ln(2)}{13} t = \ln\left(\dfrac{18}{355}\right)[/tex]
[tex]\implies t = -\dfrac{13 \ln\left(\frac{18}{355}\right)}{\ln(2)}[/tex]
[tex]\implies t = -13 \log_2\left(\dfrac{18}{355}\right) \approx \boxed{55.9}[/tex]
