What is the solution for log equations with a base of 4 and a difference of 3?

[nae99]

Homework Statement

log4 x - log4 (x-3) = 5

Homework Equations

The Attempt at a Solution

log4 x/x-3 = 5

 

[dextercioby]

OK, so it's something like

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

What do you do next ?

 

[nae99]

i think

x/ x-3 = 4^5

 

[dextercioby]

Good thinking. Next ?

 

[nae99]

x$/$ x-3 = 1024

 

[dextercioby]

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

 

[nae99]

no that's where i am stuck

 

[nae99]

i would probably do this:

x = 1024 (x-3)

 

[nae99]

??

 

[dextercioby]

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

 

[nae99]

yes now it will be
x = 1023x - 3072

 

[dextercioby]

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 ?

 

[nae99]

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

x = 1024 - 3072

 

[nae99]

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]

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

 

[nae99]

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]

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=... ?

 

[nae99]

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]

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.

 

[nae99]

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]

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?

 

[nae99]

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 don't understand how to do that

 

[Mentallic]

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

 

[nae99]

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]

Just check to see if x>3

 

[nae99]

Just check to see if x>3

and how do i go about doing that... i really don't understand that part

 

[Mentallic]

You just found that x=3072/1023, is this more than 3?

 

[nae99]

You just found that x=3072/1023, is this more than 3?

no its = 3

 

[Mentallic]

Not quite, 3*1023=3069

 

[nae99]

not quite, 3*1023=3069

-3072/1023 = -3.003