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

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Homework Statement



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


Homework Equations





The Attempt at a Solution


log(x-1)^2 + logx = logx (4)
 

Answers and Replies

  • #2


Hey there,

ok it looks like you have some exact same term on both sides to begin with, so those can just cancel out right? This leaves you with a log of something = a log of something else, that should simplify it down nice for ya.
 
  • #3
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i dont get it
 
  • #4
129
0


Hey there,

ok it looks like you have some exact same term on both sides to begin with, so those can just cancel out right? This leaves you with a log of something = a log of something else, that should simplify it down nice for ya.
so are u saying it should be:
log (2x - x)^2 = 4x
 
  • #5


Well the first step is just an algebra step, you have log (x) on both sides of the = sign, so the equation can be reduced. When that happens you are left with a log (of something) = log (of something else) when this happens you can sort of drop the log and just solve for x.
 
  • #6
3,812
92


so are u saying it should be:
log (2x - x)^2 = 4x
Noooooo, :)
What you need to do here is just first cancel out the terms [itex]\log x[/itex] on both sides.
Then shift [itex]\log 4[/itex] to the left side remaining 0 at the right side.
Now using these two identities, try to figure out the answer:-
[tex]p(\log_a b)=log_a b^p[/tex]
and
[tex]\log_a u-\log_a v=\log_a \frac{u}{v}[/tex]
 
  • #7
129
0


Well the first step is just an algebra step, you have log (x) on both sides of the = sign, so the equation can be reduced. When that happens you are left with a log (of something) = log (of something else) when this happens you can sort of drop the log and just solve for x.
log (x-1) = log4

i dont think i understand that
 
  • #8
129
0


Noooooo, :)
What you need to do here is just first cancel out the terms [itex]\log x[/itex] on both sides.
Then shift [itex]\log 4[/itex] to the left side remaining 0 at the right side.
Now using these two identities, try to figure out the answer:-
[tex]p(\log_a b)=log_a b^p[/tex]
and
[tex]\log_a u-\log_a v=\log_a \frac{u}{v}[/tex]
2log (x-1) - log 4 = 0
 
  • #9


2log (x-1) - log 4 = 0
There you go! I'd have kept log(4) on the right side and then just solve for x.
 
  • #10
3,812
92


2log (x-1) - log 4 = 0
Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave. :smile:
 
  • #11
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Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave. :smile:
log (x-1)^2 - log 4 = 0
 
  • #12
3,812
92


log (x-1)^2 - log 4 = 0
Now use the second identity. :wink:
 
  • #14
3,812
92


log (x^2-1) /4 = 0
It would be
[tex]\log \frac{(x-1)^2}{4}=0[/tex]

Whatever the base would be, when you use the antilog method here, the right hand side would become 1. So what you get is:-
[tex]\frac {(x-1)^2}{4}=1[/tex]

Now solve it further. :wink:
 
  • #15
129
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It would be
[tex]\log \frac{(x-1)^2}{4}=0[/tex]

Whatever the base would be, when you use the antilog method here, the right hand side would become 1. So what you get is:-
[tex]\frac {(x-1)^2}{4}=1[/tex]

Now solve it further. :wink:
ok so i would do this next but i dont think its right

(x^2-1) /4 = 1
 
  • #16
3,812
92


ok so i would do this next but i dont think its right

(x^2-1) /4 = 1
I don't understand how you convert (x-1)2 to (x2-1)? :confused:
That's wrong!!

Solving this equation:-
[tex]\frac {(x-1)^2}{4}=1[/tex]
we get,
[tex](x-1)^2=4[/tex]
Now try to do it.
 
  • #17
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ok, but i thought (x-1) meant (x-1) (x-1)
 
  • #18
3,812
92


ok, but i thought (x-1) meant (x-1) (x-1)
I still didn't get you :confused:
[tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]
 
  • #19
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I still didn't get you :confused:
[tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]
so i would end up with:

x^2-2x+1 = 4
 
  • #20
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and then i would have to do the quadratic equation, right
 
  • #21
3,812
92


so i would end up with:

x^2-2x+1 = 4
Yes, now proceeding further you get
[tex]x^2-2x-3=0[/tex]
Now this is quadratic equation, try to solve it. :)
 
  • #22
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Yes, now proceeding further you get
[tex]x^2-2x-3=0[/tex]
Now this is quadratic equation, try to solve it. :)
x = -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] (-2)^2 - 4*1*(-3)[itex]/[/itex] 2*1

x= -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] -4+12[itex]/[/itex] 2

how is that
 
  • #23
3,812
92


x = -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] (-2)^2 - 4*1*(-3)[itex]/[/itex] 2*1

x= -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] -4+12[itex]/[/itex] 2

how is that
What is the square of (-2)? :smile:
 
  • #24
129
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:confused:i dont know
 

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