How do you solve #(Logx)^2 + Log(x^3) + 2 = 0#?

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Answer 1 · 2016-02-14T00:11:56

#(log x)^2 +log(x^3)+2=0#

log(a^b)=blog(a)

#(log x)^2 +3log(x)+2=0#

This is one expresssion of the type ay^2 +by +c=0

Now we solve the second degree equation:

#log(x)=(-3+-sqrt(3^2-4*2))/2#

#log(x)=(-3+-1)/2#

#log(x)=-2/2 or log(x)=-4/2#

#log(x)=-1 or log(x)=-2#

#x=e^-1 or x=e^-2#