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?
Hello, RD33!
$\displaystyle \text{Solve for }x\!:\;\;e^{2x}-7e^x+12\:=\:0$
Factor: .$\displaystyle (e^x - 3)(e^x - 4) \:=\:0$
Then: .$\displaystyle \begin{Bmatrix}e^x-3 \:=\:0 & \Rightarrow & e^x \:=\:3 & \Rightarrow & x \:=\:\ln 3 \\ e^x-4 \:=\:0 & \Rightarrow & e^x \:=\:4 & \Rightarrow & x \:=\:\ln4 \end{Bmatrix}$