e^x - 6e^(-x) = 1 solve for x I think you can make 1=lne or e^0 and that helps but i'm still having trouble at the next step.
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If you multiply through by e^x you should end up with an equation you can solve
Multiply both sides by e^x to get: $\displaystyle e^{2x} - 6 = e^x$ Rearrange it into a quadratic: $\displaystyle e^{2x} - e^x - 6 = 0$ Notice that this factors into: $\displaystyle (e^x-3)(e^x+2)=0$ Can you continue?
got it, thanks alot guys!
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