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The formula A= Pe^rt, calculate to the nearest hundreth of a year how long it takes for an amount of money to double if interest is compounded continuously at 6.2%
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solve \[e^{.062t}=2\] for \(t\)
How do I reach there again...?
Do I add the on to both sides?
Ln**
A= P*e^rt You have to solve for 't'
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You need the total (A) to be twice the beginning amount so 2 = 1 * e^.062 * t to solve for t you need to take the log of both sides
But because of the e, its ln, isn't it? Like ln2 = lne^.062t
ln means natural log ln(2) = .062 * t * ln(e) 0.6931471806 = .062 * t * 1 t = 0.6931471806 / .062 t = 11.1797932355 years t= 11.18 years (rounded)
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