## rosedewittbukater 4 years ago 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.

1. rosedewittbukater

@whpalmer4 do you think you can help?

2. whpalmer4

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.

3. rosedewittbukater

Thanks so much! Especially for the detailed explanation. I really appreciate it!

4. whpalmer4

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!

5. rosedewittbukater

Yeah it's really helpful! Helps me understand it for the test coming up.