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?
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Originally Posted by MaxJasper Roots are:
x2=ln(4) How did you get that?
Let u=e^x then
solve for u then plug into u=e^x and find x.
Last edited by MaxJasper; Oct 15th 2012 at 09:07 AM.
This is what I got when I solved for u:
The log is base e^x
Am I right?
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