now, the initial amount is 900 grams, so half of that will be 450 grams.
so, how long will it be, for D(t) to turn to 450 grams?
[tex]\bf D(t)=900e^{-0.002415t}\implies 450=900e^{-0.002415t}
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\cfrac{450}{900}=e^{-0.002415t}\implies \cfrac{1}{2}=e^{-0.002415t}\\\\
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\textit{Logarithm Cancellation Rules}\\\\
log_{{ a}}{{ a}}^x\implies x\qquad \qquad
{{ a}}^{log_{{ a}}x}=x\\\\
-------------------------------\\\\[/tex]
[tex]\bf log_e\left( \frac{1}{2} \right)=log_e\left( e^{-0.002415t} \right)\implies log_e\left( \frac{1}{2} \right)=-0.002415t
\\\\\\
\cfrac{ln\left( \frac{1}{2}\right)}{-0.002415}=t\implies 287.0174661 \approx t[/tex]