# Log4 x - log4 (x-3) = 5

## Homework Statement

log4 x - log4 (x-3) = 5

## The Attempt at a Solution

log4 x/x-3 = 5

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dextercioby
Homework Helper

OK, so it's something like

$$\log_{4} \frac{x}{x-3} = 5$$

What do you do next ?

i think

x/ x-3 = 4^5

dextercioby
Homework Helper

Good thinking. Next ?

x$/$ x-3 = 1024

dextercioby
Homework Helper

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

no thats where i am stuck

i would probably do this:

x = 1024 (x-3)

??

dextercioby
Homework Helper

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

yes now it will be
x = 1023x - 3072

dextercioby
Homework Helper

I think you're missing something very simple. You've got an equation in which you must what numerical value hides under 'x'.

So $x= 1024\cdot (x-3) = 1024\cdot x - 1024\cdot 3 = 1024\cdot x - 3072$

Now subtract x from both terms of $x= 1024\cdot x - 3072$.

What do you get ?

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

x = 1024 - 3072

I think you're missing something very simple. You've got an equation in which you must what numerical value hides under 'x'.

So $x= 1024\cdot (x-3) = 1024\cdot x - 1024\cdot 3 = 1024\cdot x - 3072$

Now subtract x from both terms of $x= 1024\cdot x - 3072$.

What do you get ?
x = 1024 - 3072

Mentallic
Homework Helper

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

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

Mentallic
Homework Helper

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=... ?

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

Mentallic
Homework Helper

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.

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

Mentallic
Homework Helper

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?

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

Mentallic
Homework Helper

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

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

Mentallic
Homework Helper

Just check to see if x>3