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suppose a 1000$ was invested in an account with interest compounded continously. If at the end of 5 years the account holds 2000$. what was the interest rate on the account ? i just dont know how to solve for r
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I have already shown you the method. See your previous question.
but how do i replace for r ?
this is compounded continuously...
The formula for continuous compounding is \[A=P \times e ^{rt}\] \[2000=1000\times e ^{5r}\] Taking logs of both sides: \[\ln 2000=\ln 1000+5r\] \[5r=\ln 2000-\ln 1000\] \[r=\frac{\ln 2000-\ln 1000}{5}\]
0.06 ?
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or 0.14?
\[r=\frac{\ln 2000-\ln 1000}{5}=0.1386\]
thanks
You're welcome :)
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