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

[Saitama]

is it -4

Nooo, it's 4.
(-2)²=(-2)(-2)=4

 

[nae99]

oh ok

 

[Saitama]

oh ok

Now what's the answer?

 

[nae99]

Now what's the answer?

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

 

[Saitama]

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.
Your both the answers are wrong. Correct them.

 

[nae99]

What are you doing?
6/2=3 and 2-2=0.
Your both the answers are wrong. Correct them.

OMG, such simple mistakes huh, ok

x= 6/2 = 3

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

 

[Mentallic]

That's too much unnecessary work. From,

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

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

 

[Saitama]

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

 

[nae99]

thanks much