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

Use a dummy variable to put the equation into a more familiar form, then substitute back in to get to $x = {e}^{\frac{1}{2}} \mathmr{and} {e}^{- 1}$

#### Explanation:

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

In one sense, this is a very complex and complicated equation. But in another sense, we can change the form of the equation to a simple trinomial by substituting a dummy variable, say A, for $\ln x$. So we can:

$A = \ln x$

therefore:

$2 {A}^{2} + A = 1$

now let's solve for A:

$2 {A}^{2} + A - 1 = 0$

$\left(2 A - 1\right) \left(A + 1\right) = 0$

So $A = \frac{1}{2} \mathmr{and} - 1$

Which means that:

$\ln x = \frac{1}{2} \mathmr{and} - 1$

$x = {e}^{\frac{1}{2}} \mathmr{and} {e}^{- 1}$