Logarithmical inequality

Okay, so I almost solved that one

I found x² - 2x -2 < 0
so x = 1 + sqrt(3)
or x = 1 - sqrt(3)

since I was looking for the values that returned a number less than 0, I thought that it would be
x > 1 - sqrt(3) and x < 1 + sqrt(3)

but the answer sheet says it's 2 and 1 + sqrt(3)

WHERE DID THAT 2 CAME FROM???

 

log(2,x)+log(1/2,(x+1))>log(2,(x-2)).......change to base 10

log(x)/log(2)+log(x+1)/log(1/2)>log(x-2)/log(2).....since log(1/2)=log(1)-log(2)=0-log(2)=log(2) we have

log(x)/log(2)-log(x+1)/log(2)>log(x-2)/log(2).........since all denominators same

log(x)-log(x+1)>log(x-2)

log(x/(x+1))-log(x-2)>0

log((x/(x+1))/(x-2))>0

log(x/(x^2 - x - 2))>0

(x^2 - 2x - 2)/((x - 2) (x + 1))<0

Real solutions:

-1<x<1 - sqrt(3)
2<x<1 + sqrt(3)

 

thank you very much, ma'am!!
what do you think about my answer?
instead of changing them to base 10, I changed the log(1/2,(x+1)) to base 2, which gave me "- log(2,(x+1))" do you think that would work?

 

that is same changing them to base 10, you just made shortcut

 

I just don't get that part right here

"log(x/(x^2 - x - 2))>0

(x^2 - 2x - 2)/((x - 2) (x + 1))<0"

do you mind explaining that a little bit?

I mean, I can see how we get to "log(x/(x^2 - x - 2))>0 ", I could do that in base 2

but how x/(x-2)(x+1) becomes (x^2 - 2x - 2)/((x - 2) (x + 1)) ? T_T

 

log(2,x)+log(1/2,(x+1))>log(2,(x-2))

log(2,x)-log(2,(x-2))-log(2,(x+1))>0

log(2,x/(x-2))-log(2,(x+1))>0

log(2,x/(x-2))>log(2,(x+1))

we know that x>2 (denominator cannot be zero)

now apply log rule:

x/(x-2)> x+1 and x+1>0

x> (x-2)(x+1 )
x> x^2 - x - 2
0> x^2 - x - 2-x ............we just flip the sides

x^2 - 2x - 2 <0-> x = 1 - sqrt(3) or x = 1 + sqrt(3)->

and x+1>0 -> and x>-1

so, solution is:
-1<x<1 - sqrt(3)
2<x<1 + sqrt(3)

 

aaaahh, I see! I was forgetting to check the conditions of existence every time (making sure I'm not dividing by zero/ that the argument of the logarithm itself is greater than 0)
now everything makes sense