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And t is the number of years.

ah yes...Pert :)

so would this A(t)=5000xe^.069(30) be correct.. ?

Got it :]

Yes. \(A(t) = 5000 * e^{.069*30} = 5000 * e^{2.07} = 5000 * 7.9248 = 39624.11\)

Thanksss ! :D

Not quite the same. But similiar. You were probably given an equation for exponential decay?

yuppp pretty much ! =)

Well ok, exponential decay has the form
\(A(t) = A_0*2^{\frac{t}{t_{halflife}}}\)

err The exponent on the 2 should be \[\frac{-t}{t_{halflife}}\]

And \(A_0\) is again the initial amount.