# How do you solve (e^x+e^-x)/2=3?

Jun 7, 2016

x≈1.82

#### Explanation:

From the given equation, we derive that

${e}^{x} - {e}^{-} x = 6$.

Note that ${e}^{-} x = \frac{1}{{e}^{x}}$ and the expression can be simplfied as

${e}^{x} - \frac{1}{{e}^{x}} = 6$
${\left({e}^{x}\right)}^{2} - 6 \left({e}^{x}\right) - 1 = 0$

Using the quadratic formula, we find that

${e}^{x} = \frac{6 \pm \sqrt{36 - 4 \left(- 1\right)}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2 \sqrt{10}}{2} = 3 \pm \sqrt{10}$

Thus, e^x = 3+sqrt(10), x = ln(3+sqrt(10)) = 1.8184 ≈ 1.82 (3sf)

or

${e}^{x} = 3 - \sqrt{10} < 0$ (no solutions)