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that the answer

yes
well you have to plug that k back into the top equation

so it is\[y=Ae-1n2/15^t\]

is that right

correct :)

\[y=Ae-1n2/15^t\]

is that a subtractions sign ???
\[y = Ae ^{-\frac{\ln 2}{15}t} \]

THANKS