SOLUTION: find the interest rate needed for $6500 to grow to $10400 in 3 years if interest is continously compounded.

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Question 612930: find the interest rate needed for $6500 to grow to $10400 in 3 years if interest is continously compounded.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A=Pe%5E%28rt%29 Start with the continuous compounding formula.


10400=6500%2Ae%5E%28r%2A3%29 Plug in A=10400, P=6500, and t=3.


10400%2F6500=e%5E%28r%2A3%29 Divide both sides by 6500.


1.6=e%5E%28r%2A3%29 Evaluate 10400%2F6500 to get 1.6.


ln%281.6%29=ln%28e%5E%28r%2A3%29%29 Take the natural log of both sides.


ln%281.6%29=r%2A3%2Aln%28e%29 Pull down the exponent using the identity ln%28x%5Ey%29=y%2Aln%28x%29%29.


ln%281.6%29=r%2A3%2A1 Evaluate the natural log of 'e' to get 1.


ln%281.6%29=r%2A3 Multiply and simplify.


0.470003629245736=r%2A3 Evaluate the natural log of 1.6 to get 0.470003629245736 (this value is approximate).


0.470003629245736%2F3=r Divide both sides by 3 to isolate 'r'.


0.156667876415245=r Evaluate 0.470003629245736%2F3 to get 0.156667876415245.


r=0.156667876415245 Flip the equation.


r=0.1567 Round to the nearest ten-thousandth.


So the interest rate is roughly 15.67%