How do you solve $2e^(2x)-9e^x -5=0$ ?

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Answer 1 · 2016-05-10T12:03:57

$x=ln 5.$

This is a quadratic in $e^x$ . The roots are $e^x=(9+-sqrt(121))/4$

$=5, 1/2$ .

As $e^x>=0$ , negative root is inadmissible..

Inverting, x= ln 5. .

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