# How do solve 3e^x=2e-x+4?

Nov 8, 2015

If you mean $3 {e}^{x} = 2 {e}^{- x} + 4$, then rearrange as a quadratic in ${e}^{x}$, solve and take logs to find:

$x = \ln \left(\frac{2 + \sqrt{10}}{3}\right)$

#### Explanation:

Assuming you meant $3 {e}^{x} = 2 {e}^{- x} + 4$, first multiply by ${e}^{x}$ to get:

$3 {\left({e}^{x}\right)}^{2} = 2 + 4 \left({e}^{x}\right)$

Subtract the right hand side from the left to get:

$3 {\left({e}^{x}\right)}^{2} - 4 \left({e}^{x}\right) - 2 = 0$

Using the quadratic formula, we get:

${e}^{x} = \frac{4 \pm \sqrt{{4}^{2} - \left(4 \times 3 \times - 2\right)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 24}}{6}$

$= \frac{4 + = \sqrt{40}}{6} = \frac{4 \pm 2 \sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}$

Now $\sqrt{10} > 2$, so to get a positive value for ${e}^{x}$ (and hence a Real value for $x$) we need to choose the $+$ sign here to find:

${e}^{x} = \frac{2 + \sqrt{10}}{3}$

and hence:

$x = \ln \left(\frac{2 + \sqrt{10}}{3}\right)$