How do you solve $7 e^{x} = 9 - e^{- x}$ $7e^{x} = 9- e^{- x}$ ?

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How do you solve $7 e^{x} = 9 - e^{- x}$ $7e^{x} = 9- e^{- x}$ ?

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Answer 1 · 2016-11-13T22:06:51

Two solutions
$x_{1} = ln (\frac{9 - \sqrt[2]{53}}{14})$ $x_1=ln((9-root2(53))/14)$
$x_{2} = ln (\frac{9 + \sqrt[2]{53}}{14})$ $x_2=ln((9+root2(53))/14)$

Explanation:

Let's rewrite the equation
$7 e^{x} - 9 + \frac{1}{e^{x}} = 0$ $7e^x-9+1/e^x=0$
We can multiply both sides for $e^{x}$ $e^x$ so that we get

$7 e^{2 x} - 9 e^{x} + 1 = 0$ $7e^(2x)-9e^x+1=0$
this can be solved as an ordinary second degree eqaution

$e^{x} = \frac{9 \pm \sqrt[2]{81 - 28}}{14}$ $e^x=(9+-root2(81-28))/14$

and consequently

$x = ln (\frac{9 \pm \sqrt[2]{53}}{14})$ $x=ln((9+-root2(53))/14)$

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How do you solve $7 e^{x} = 9 - e^{- x}$ $7e^{x} = 9- e^{- x}$ ?

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