Solve exponential inequation $(e^x)^2-e^x.e^1 - e^x + e < 0$

Question

I have the equation $e^{2x}-e^{x+1} - e^x + e < 0$. I solved that equation until here:

$(e^x)^2-e^x.e^1 - e^x + e < 0$

$y=e^x$

$(y)^2-y.e^1 - y + e < 0$

$y^2-y.e - y + e < 0$

But from here i am not able to solve through a quadratic equation. How can i solve that?

Answer 1

It's $$e^x(e^x-e)-(e^x-e)<0$$ or $$(e^x-e)(e^x-1)<0$$ or $$1<e^x<e$$ or $$0<x<1.$$ Your quadratic inequality we can solve by the same way: $$y^2-ey-y+e<0$$ or $$y(y-e)-(y-e)<0$$ or $$(y-1)(y-e)<0$$ or $$1<y<e.$$