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

1 Answer

Answer:

Use a dummy variable to put the equation into a more familiar form, then substitute back in to get to #x=e^(1/2) and e^(-1)#

Explanation:

Let's start with the original:

#2(lnx)^2+lnx=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 #lnx#. So we can:

#A=lnx#

therefore:

#2A^2+A=1#

now let's solve for A:

#2A^2+A-1=0#

#(2A-1)(A+1)=0#

So #A=1/2 and -1#

Which means that:

#lnx=1/2 and -1#

#x=e^(1/2) and e^(-1)#