How do you solve #e^[2 ln x – ln (x^2 + x - 3)] = 1#?

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Answer 1 · 2016-05-06T10:55:37

#x = 3#

Use #m ln n=ln m^n and e^log a=a#

The given equation is

#e^(ln x^2-ln(x^2+x-3))=1#

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

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

#x^2=x^2+x-3#

#x=3#