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rosedewittbukater
Algebra 2 help and explanation? 2 problems please. Thanks! 1) 4e^(x-1) = 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.
@whpalmer4 do you think you can help?
I think I can, I think I can :-) \[4e^{x-1} = 64\]Divide both sides by 4\[e^{x-1}=16\]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 \[\ln{e^{x-1}}=x-1=\ln{16}\]\[x=1+\ln{16} \approx 3.77\]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.
Thanks so much! Especially for the detailed explanation. I really appreciate it!
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!
Yeah it's really helpful! Helps me understand it for the test coming up.