# How do you solve e^x + e^(-x) = 3?

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#### Explanation:

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14
Jan 1, 2016

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

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

#### Explanation:

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

Then the equation becomes:

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

Multiplying both sides by $t$ we get:

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

Subtract $3 t$ from both sides to get:

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

Use the quadratic formula to find roots:

$t = \frac{3 \pm \sqrt{5}}{2}$

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

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

${e}^{x} = \frac{3 \pm \sqrt{5}}{2}$

Taking natural logs of both sides we find:

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

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