How do you solve #(ln(x))^2 + ln(x) - 6 = 0#?

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Answer 1 · 2018-06-05T15:35:03

#x=e^(-3)=1/(e^3) or x=e^2#

We know that,

#(1)log_a y=X <=>y=a^X, AAa in RR^+ -{1},yinRR^+,X inRR#

#(2)color(red)(ln(y)=X<=>y=e^X ),where,yinRR^+ andcolor(red)( X inRR#

Here,

#(ln(x))^2+ln(x)-6=0#

Let,

#ln(x)=m#

So,

#m^2+m-6=0#

#m^2+3m-2m-6=0#

#m(m+3)-2(m+3)=0#

#(m+3)(m-2)=0#

#m+3=0 or m-2=0#

#m=-3 or m=2#

Subst. back, #m=ln(x)#

#:.lnx=-3 or lnx=2,where,color(red)( (-3) and 2inRR#

#=>x=e^(-3) or x=e^2#