Question

How do you solve `2(lnx)^2 + lnx -1 = 0`?

Answer 1

`x=e^(1/2)color(white)("XXX")orcolor(white)("XXX")x=e^(-1)`

Explanation:

Let `a=ln(x)`

Then
`color(white)("XXX")2(ln(x))^2+ln(x)-1=0`
`color(white)("XXX")hArr 2a^2+a-1=0`
which can be factored as
`color(white)("XXX")(2a-1)(a+1)=0`

Which implies
`color(white)("XXX")a=1/2 or a=-1`

and since `a=ln(x)`
`color(white)("XXX")ln(x)=1/2 or ln(x)=-1`

`color(white)("XXX")x=e^(1/2) or x=e^(-1)`