Help finding x in this equation

Note the reason for not be able to take a negative log.......

If we had

log (-2)     ......this would imply that  there is some "x" such that

10^x  = -2

But notice that 10 raised to  any  "real" power is always positive....therefore.......there is no "real" power "x"   that we can raise 10 to that will result in a negative.......

Solve for x:
log(x)+log(x+1) = log(2)

log(x)+log(x+1) = log(x (x+1)):
log(x (x+1)) = log(2)

Cancel logarithms by taking exp of both sides:
x (x+1) = 2

Expand out terms of the left hand side:
x^2+x = 2

Add 1/4 to both sides:
x^2+x+1/4 = 9/4

Write the left hand side as a square:
(x+1/2)^2 = 9/4

Take the square root of both sides:
x+1/2 = 3/2 or x+1/2 = -3/2

Subtract 1/2 from both sides:
x = 1 or x+1/2 = -3/2

Subtract 1/2 from both sides:
x = 1 or x = -2

log(x)+log(x+1) ⇒ log(-2)+log(1-2) = 2 i π+log(2) ≈ 0.693147+6.28319 i
log(2) ⇒ log(2) ≈ 0.693147:
So this solution is incorrect

log(x)+log(x+1) ⇒ log(1)+log(1+1) = log(2) ≈ 0.693147
log(2) ⇒ log(2) ≈ 0.693147:
So this solution is correct

The solution is:
Answer: |x = 1

logx+log(x+1)=log2    

Remember that if we have log a + log b, we can write this as log [a * b]........so....

log x + log (x + 1)   =  log [ x (x + 1) ]  =  log [x ^2 + x]

So

log [ x^2 + x  ]    = log 2         we can just forget the logs and solve this equation :

x^2 + x  = 2      subtract 2 from both sides

x^2 + x  - 2  = 0         factor

(x -1) (x + 2)   = 0       and setting each factor to 0, x = 1  or x = -2

Since we can't take the log of a negative number, x = - 2 is not a solution

Check  x = 1

log (1) + log (1 + 1) =  log 2 ?

0 + log 2   = log 2

log 2 = log 2       OK!!!

I'll write here the solution included in my book:

logx + log(x+1) = log2

logx(x+1)= log2

logx(x+1)-log2=0

and then

log(x(x+1)/2)=0  (can someone explain me this step plz)

x(x+1)/2= 10exp0=1  (I feel this is some basic log law but I can't remember what it is exactly, help haha)

Then they multiply each side by 2 to get rid of the /2:

x^2+x-2=0  (why do they end up with "x^2" and "-2" when they just get rid of the /2?)

then they say "logx(x+1)=log2" means that "x(x+1)=2" and that "x^2+x-2=0"

Then they find rationnal roots for "x^2+x-2=0" which are "x=1" and "x=-2" (that's fine for me)

Seems that "x=1" makes sense because "log1+ log(1+1)=log2"

but that "x=-2" because " log of a negative number is undefined" (can someone refresh me on that).

Thanks a lot guys!