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Solve for x.

ln(x-1)+ln(x+2)=1

(Does it equal a rational number?)

 Feb 12, 2019
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\(ln(x-1)+ln(x+2)=1\\ ln[(x-1)(x+2)]=1\\ e^{ln[(x-1)(x+2)]}=e^1\\ x^2+x-2=e\\ x^2+x-(2+e)=0\\\)

\(x^2+x-(2+e)=0\\ x=\frac{-1\pm\sqrt{1+4(2+e)}}{2}\\ x=\frac{-1\pm\sqrt{1+8+4e}}{2}\\ x=\frac{-1\pm\sqrt{9+4e}}{2}\\ \)

 

Those are not equal or rational solutions.

 Feb 12, 2019

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