How do you solve $Ln (x-1) + ln (x+2) = 1$ ?

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Answer 1 · 2016-09-22T15:55:41

$x=1.729$

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

$hArrln(x-1)(x-2)=1$ or

$(x-1)(x+2)=e$ or

$x^2+x-2-e=0$ and using quadratic fomula $(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-1+-sqrt(1^2-4xx1xx(-2-e)))/2$

= $(-1+-sqrt(1+8+4e))/2$

= $(-1+-sqrt(9+4e))/2$

= $(-1+-sqrt(9+4xx2.7183))/2$

= $(-1+-sqrt(9+10.8732))/2$

= $(-1+-sqrt(19.8732))/2$

= $(-1+-4.4579)/2$

i.e. $x=1.729$ or $x=-2.729$

But $x$ cannot have a negative value, hence $x=-2.729$

How do you solve Ln (x-1) + ln (x+2) = 1 Ln (x-1) + ln (x+2) = 1 ?