Answer
See below.
Work Step by Step
We know that $0.5x=xe^{k\cdot5730}\\0.5=e^{5730k}\\5730k=\ln0.5\\k=\frac{\ln0.5}{5730}\approx-0.000120968$
Thus $0.3=e^{-0.000120968\cdot t}\\-0.000120968t=\ln0.3\\t=\frac{\ln0.3}{-0.000120968}\approx9953$
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.