# How do you solve ln(2x+1) = 2 - ln(x)?

Jun 27, 2018

$\ln \left(2 x + 1\right) = 2 - \ln \left(x\right)$

${e}^{\ln \left(2 x + 1\right)} = {e}^{2 - \ln \left(x\right)}$

$2 x + 1 = {e}^{2} / x$

$2 {x}^{2} + x = {e}^{2}$

$2 {x}^{2} + x - {e}^{2} = 0$

$x = \frac{- 1 \pm \sqrt{\left(1 + 8 \left({e}^{2}\right)\right)}}{4}$

$2 \cdot \frac{- 1 - \sqrt{\left(1 + 8 \left({e}^{2}\right)\right)}}{4} + 1 < 0$

Therefore $\ln \left(2 \cdot \frac{- 1 - \sqrt{\left(1 + 8 \left({e}^{2}\right)\right)}}{4} + 1\right)$ is not defined and not a valid input for x.

This means $x = \frac{- 1 + \sqrt{\left(1 + 8 \left({e}^{2}\right)\right)}}{4}$