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how many years will it take $2 million to grow to $5 million with an annual interest rate of 7%
 Aug 21, 2013

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 #1
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2*(1+0.07)^x = 5 [/2]
(1+0.07)^x = 5/2 [log_1.07]
x = log_1.07(5/2) = 13,xxx

You could also use e:
(1+0.07)^x = (1+7/100)^x = e^(7x/100)

e^(7x/100) = 5/2 [ln]
7x/100 = ln(5/2)
=> x = ln(5/2)*100/7 = 13,xxx
 Aug 22, 2013

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