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

Nov 30, 2015

$x = \frac{2}{e}$

#### Explanation:

We will use the following:

• $\ln \left({a}^{x}\right) = x \ln \left(a\right)$
• $\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$
• ${e}^{\ln} \left(a\right) = a$

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

$\implies \ln \left({x}^{2}\right) + 1 = \ln \left(2 x\right)$ (by the first property above)

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

$\implies \ln \left({x}^{2} / \left(2 x\right)\right) = - 1$ (by the second property above)

$\implies \ln \left(\frac{x}{2}\right) = - 1$

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

$\implies \frac{x}{2} = \frac{1}{e}$ (by the third property above)

$\therefore x = \frac{2}{e}$