- #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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- Thread starter nae99
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- #1

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log4 x - log4 (x-3) = 5

log4 x/x-3 = 5

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- #2

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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

- #4

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Good thinking. Next ?

- #5

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x[itex]/[/itex] x-3 = 1024

- #6

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So it's very easy to find the x, right ?

- #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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??

- #10

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So it's like 3072 = 1023 x. Do you agree ?

- #11

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yes now it will be

x = 1023x - 3072

- #12

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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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x = 1024 - 3072

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 ?

- #15

Mentallic

Homework Helper

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What?ok, so if i am going to subtract x from both terms, i would end up with:

x = 1024 - 3072

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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as u can see i am not good at this, ok here goesWhat?

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

1024x-x=3072

1023x = 3072

- #17

Mentallic

Homework Helper

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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.as u can see i am not good at this, ok here goes

Yes and now? If I asked you to solve for x why haven't you given us x=... ?1024x-x=3072

1023x = 3072

- #18

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x = 3072-1023Well 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 = 2049

- #19

Mentallic

Homework Helper

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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.x = 3072-1023

x = 2049

- #20

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so i would divide both sides by 1023No, 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.

1023x = 3072

1023x/1023 = 3072/1023

- #21

Mentallic

Homework Helper

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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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i dont understand how to do that

- #23

Mentallic

Homework Helper

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You need to be less vague. What don't you understand?

- #24

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how to check to see if the solutions are valid in the original questionYou need to be less vague. What don't you understand?

- #25

Mentallic

Homework Helper

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Just check to see if x>3

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