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Algebra 2 help and explanation? 2 problems please. Thanks!
1) 4e^(x1) = 64 Solve the equation.
2) An initial investment of $350 is worth $429.20 after six years of continuous compounding. Find the annual interest rate.
 one year ago
 one year ago
Algebra 2 help and explanation? 2 problems please. Thanks! 1) 4e^(x1) = 64 Solve the equation. 2) An initial investment of $350 is worth $429.20 after six years of continuous compounding. Find the annual interest rate.
 one year ago
 one year ago

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rosedewittbukaterBest ResponseYou've already chosen the best response.0
@whpalmer4 do you think you can help?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I think I can, I think I can :) \[4e^{x1} = 64\]Divide both sides by 4\[e^{x1}=16\]Take the natural log of both sides to get rid of the e^, remembering that \[\ln{e^a}=a\] because the log of a number is simply the power that the base of the log must be raised to to give the number \[\ln{e^{x1}}=x1=\ln{16}\]\[x=1+\ln{16} \approx 3.77\]We could also notice that 16 = 4^2 and write \[1+\ln{4^2} = 1+4\ln{2}\] Continuous compounding with interest rate r (written as a decimal, not a percentage) gives \[FV=PVe^{rt}\] where FV is future value and PV is present value. Here we have FV=429.20, PV=350, t = 6, and we need to find r. \[FV=PVe^{rt}\]Divide by PV and take log of both sides\[\ln{(\frac{FV}{PV})} = rt\]Divide by t to get r in terms of FV,PV,t\[r=\frac{1}{t}\ln{(\frac{FV}{PV})}\]and now it is just a matter of pushing buttons on the calculator. Remember that the rate you get is a decimal, and to convert it to a percentage you need to multiply by 100.
 one year ago

rosedewittbukaterBest ResponseYou've already chosen the best response.0
Thanks so much! Especially for the detailed explanation. I really appreciate it!
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I figure there's no point in simply giving answers, but if I show how I got them, hopefully you'll be able to see how to do them yourself!
 one year ago

rosedewittbukaterBest ResponseYou've already chosen the best response.0
Yeah it's really helpful! Helps me understand it for the test coming up.
 one year ago
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