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

1 Answer
Feb 15, 2016

Answer:

#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)#