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

  • Thread starter nae99
  • Start date
  • #26
3,812
92


is it -4
Nooo, it's 4.
(-2)2=(-2)(-2)=4
 
  • #27
129
0


oh ok
 
  • #29
129
0


Now what's the answer?
x= -(-2) ± √ 4+12/ 2

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4
OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1
 
  • #30
3,812
92


x= -(-2) ± √ 4+12/ 2

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4
OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1
What are you doing?
6/2=3 and 2-2=0.
Your both the answers are wrong. Correct them.
 
  • #31
129
0


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
 
  • #32
Mentallic
Homework Helper
3,798
94


That's too much unnecessary work. From,

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).
 
  • #33
3,812
92


That's too much unnecessary work. From,

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).
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
 
  • #34
129
0


thanks much
 

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