e^(2x)-7e^x+12=0 had two solutions i got this (7(plus minus)sqrt(7^2-4*e*12))/2e now i replaced e with 2.718 because that is the value of e and I got 1.591 and 0.984. Is this hat I what's suppose to do?
Follow Math Help Forum on Facebook and Google+
Roots are: x1=ln(3) x2=ln(4)
Originally Posted by MaxJasper Roots are: x1=ln(3) x2=ln(4) How did you get that?
Let u=e^x then u^2-7u+12=0 solve for u then plug into u=e^x and find x.
Last edited by MaxJasper; Oct 15th 2012 at 08:07 AM.
This is what I got when I solved for u: U= (loge^x(-12)-2)/-7 The log is base e^x Am I right?
Hello, RD33! Factor: . Then: .
View Tag Cloud