Log4 x - log4 (x-3) = 5

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  • #1
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Homework Statement



log4 x - log4 (x-3) = 5

Homework Equations





The Attempt at a Solution



log4 x/x-3 = 5
 
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Answers and Replies

  • #2
dextercioby
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OK, so it's something like

[tex] \log_{4} \frac{x}{x-3} = 5 [/tex]

What do you do next ?
 
  • #3
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i think

x/ x-3 = 4^5
 
  • #5
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x[itex]/[/itex] x-3 = 1024
 
  • #7
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no thats where i am stuck
 
  • #8
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i would probably do this:

x = 1024 (x-3)
 
  • #9
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:confused:??
 
  • #11
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yes now it will be
x = 1023x - 3072
 
  • #12
dextercioby
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I think you're missing something very simple. You've got an equation in which you must what numerical value hides under 'x'.

So [itex] x= 1024\cdot (x-3) = 1024\cdot x - 1024\cdot 3 = 1024\cdot x - 3072 [/itex]

Now subtract x from both terms of [itex] x= 1024\cdot x - 3072 [/itex].

What do you get ?
 
  • #13
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ok, so if i am going to subtract x from both terms, i would end up with:

x = 1024 - 3072
 
  • #14
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I think you're missing something very simple. You've got an equation in which you must what numerical value hides under 'x'.

So [itex] x= 1024\cdot (x-3) = 1024\cdot x - 1024\cdot 3 = 1024\cdot x - 3072 [/itex]

Now subtract x from both terms of [itex] x= 1024\cdot x - 3072 [/itex].

What do you get ?
x = 1024 - 3072
 
  • #15
Mentallic
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ok, so if i am going to subtract x from both terms, i would end up with:

x = 1024 - 3072
What?

Don't you know basic algebra? Solve for x in 3x=x+2, now apply the same idea to solve for x in x=1024x-3072
 
  • #16
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What?

Don't you know basic algebra? Solve for x in 3x=x+2, now apply the same idea to solve for x in x=1024x-3072
as u can see i am not good at this, ok here goes

1024x-x=3072
1023x = 3072
 
  • #17
Mentallic
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as u can see i am not good at this, ok here goes
Well then you need to go back and cover basic algebra again. You can't afford to lose that many marks on logarithms just because you don't know your algebra.

1024x-x=3072
1023x = 3072
Yes and now? If I asked you to solve for x why haven't you given us x=... ?
 
  • #18
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Well then you need to go back and cover basic algebra again. You can't afford to lose that many marks on logarithms just because you don't know your algebra.



Yes and now? If I asked you to solve for x why haven't you given us x=... ?
x = 3072-1023
x = 2049
 
  • #19
Mentallic
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x = 3072-1023
x = 2049
No, it's not 1023+x=3072, it's 1023x=3072. You really need to go back and catch up on what you've missed out on.
 
  • #20
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No, it's not 1023+x=3072, it's 1023x=3072. You really need to go back and catch up on what you've missed out on.
so i would divide both sides by 1023

1023x = 3072

1023x/1023 = 3072/1023
 
  • #21
Mentallic
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Yes, now just go back and check to see if your solutions are valid in the original question. That is, everything that you take the logarithm of needs to be greater than zero, so x>0 and x-3>0, therefore x>3. Since both of these need to be satisfied, we just consider x>3. Is your solution valid?
 
  • #22
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Yes, now just go back and check to see if your solutions are valid in the original question. That is, everything that you take the logarithm of needs to be greater than zero, so x>0 and x-3>0, therefore x>3. Since both of these need to be satisfied, we just consider x>3. Is your solution valid?
i dont understand how to do that
 
  • #23
Mentallic
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You need to be less vague. What don't you understand?
 
  • #24
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You need to be less vague. What don't you understand?
how to check to see if the solutions are valid in the original question
 
  • #25
Mentallic
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Just check to see if x>3
 

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