How do you solve $e^{x} + e^{- x} = 3$ $e^x + e^(-x) = 3$ ?

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How do you solve $e^{x} + e^{- x} = 3$ $e^x + e^(-x) = 3$ ?

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Answer 1 · 2016-01-01T18:02:37

Express as a quadratic in $t = e^{x}$ $t = e^x$ , solve and take logs to find:

$x = ln (\frac{3 \pm \sqrt{5}}{2}) = \pm ln (\frac{3 + \sqrt{5}}{2})$ $x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)$

Explanation:

Let $t = e^{x}$ $t = e^x$ .

Then the equation becomes:

$t + \frac{1}{t} = 3$ $t + 1/t = 3$

Multiplying both sides by $t$ $t$ we get:

$t^{2} + 1 = 3 t$ $t^2+1 = 3t$

Subtract $3 t$ $3t$ from both sides to get:

$t^{2} - 3 t + 1 = 0$ $t^2-3t+1 = 0$

Use the quadratic formula to find roots:

$t = \frac{3 \pm \sqrt{5}}{2}$ $t = (3+-sqrt(5))/2$

Note that due to the symmetry of the equation $t + \frac{1}{t} = 3$ $t+1/t = 3$ in $t$ $t$ and $\frac{1}{t}$ $1/t$ , these two values are actually reciprocals of one another.

Now $t = e^{x}$ $t = e^x$ , so:

$e^{x} = \frac{3 \pm \sqrt{5}}{2}$ $e^x = (3+-sqrt(5))/2$

Taking natural logs of both sides we find:

$x = ln (\frac{3 \pm \sqrt{5}}{2}) = \pm ln (\frac{3 + \sqrt{5}}{2})$ $x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)$

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