# 2log(x-1) + logx = logx + log4

is it -4
Nooo, it's 4.
(-2)2=(-2)(-2)=4

oh ok

oh ok

x= -(-2) ± √ 4+12/ 2

x = -(2) $\pm$$\sqrt{}$ 16 $/$ 2

x = -(-2) $\pm$ 4 $/$ 2

x = -(-2) + 4 $/$ 2

x = 6$/$ 2

x= 4
OR

x = -(-2) - 2 $/$ 2

x = -2$/$ 2

x = -1

x= -(-2) ± √ 4+12/ 2

x = -(2) $\pm$$\sqrt{}$ 16 $/$ 2

x = -(-2) $\pm$ 4 $/$ 2

x = -(-2) + 4 $/$ 2

x = 6$/$ 2

x= 4
OR

x = -(-2) - 2 $/$ 2

x = -2$/$ 2

x = -1
What are you doing?
6/2=3 and 2-2=0.

What are you doing?
6/2=3 and 2-2=0.
OMG, such simple mistakes huh, ok

x= 6/2 = 3

x= -(-2) - 2/ 2 = 0/2 x= 0

Mentallic
Homework Helper

That's too much unnecessary work. From,

$$(x-1)^2=4$$

Take the square root of both sides (remember the $\pm$).

That's too much unnecessary work. From,

$$(x-1)^2=4$$

Take the square root of both sides (remember the $\pm$).
You're right Mentallic but nae99 is doing such silly mistakes here in this thread.

@nae99- take care, since you need to consider this case:- x-1>0

thanks much