Solve the inequation

[DanieldeLucena]

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[Deleted member 4993]

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\(\displaystyle e^x - e^{x-1} - e^x + e < 0\)

\(\displaystyle - e^{x-1} + e < 0\)

\(\displaystyle e^{x-1} > e\)

Now continue....

 

[DanieldeLucena]

LOL man sorry about that I mean e^2x and e^x+1... like it ...
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My fault ...

 

[Deleted member 4993]

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\(\displaystyle e^{2x} - e^{x+1} - e^x + e < 0\)

Substitute

e[sup:u3o35p6r]x[/sup:u3o35p6r] = u

u[sup:u3o35p6r]2[/sup:u3o35p6r] - u(e+1) + e < 0

Now continue....

 

[DanieldeLucena]

Man I don't knowing how to continue, look that ...
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I really don't know how to solve this, it's too difficult to me ...
Can anyone solve this complete ???

 

[DanieldeLucena]

LoL thx I solved it 0<x<1

 

[Denis]

LoL thx I solved it 0<x<1

Is LoL a mathematical symbol?

 

[Deleted member 4993]

\(\displaystyle e^{2x} - e^{x+1} - e^x + e < 0\)

Substitute

e[sup:wk8wcwfb]x[/sup:wk8wcwfb] = u

u[sup:wk8wcwfb]2[/sup:wk8wcwfb] - u(e+1) + e < 0

(u-1)(u-e) < 0

f(u) < 0

1 < u < e

1 < e[sup:wk8wcwfb]x[/sup:wk8wcwfb] < e

0 < x < 1

finished