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

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

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

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

$x^2=x^2+x-3$

$x=3$

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