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

1. Jul 7, 2011

### nae99

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution
log(x-1)^2 + logx = logx (4)

2. Jul 7, 2011

### Omega_Prime

Re: logs

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. Jul 7, 2011

### nae99

Re: logs

i dont get it

4. Jul 7, 2011

### nae99

Re: logs

so are u saying it should be:
log (2x - x)^2 = 4x

5. Jul 7, 2011

### Omega_Prime

Re: logs

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. Jul 7, 2011

### Saitama

Re: logs

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

7. Jul 7, 2011

### nae99

Re: logs

log (x-1) = log4

i dont think i understand that

8. Jul 7, 2011

### nae99

Re: logs

2log (x-1) - log 4 = 0

9. Jul 7, 2011

### Omega_Prime

Re: logs

There you go! I'd have kept log(4) on the right side and then just solve for x.

10. Jul 7, 2011

### Saitama

Re: logs

Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave.

11. Jul 7, 2011

### nae99

Re: logs

log (x-1)^2 - log 4 = 0

12. Jul 7, 2011

### Saitama

Re: logs

Now use the second identity.

13. Jul 7, 2011

### nae99

Re: logs

log (x^2-1) /4 = 0

14. Jul 7, 2011

### Saitama

Re: logs

It would be
$$\log \frac{(x-1)^2}{4}=0$$

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

Now solve it further.

15. Jul 7, 2011

### nae99

Re: logs

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

(x^2-1) /4 = 1

16. Jul 7, 2011

### Saitama

Re: logs

I don't understand how you convert (x-1)2 to (x2-1)?
That's wrong!!

Solving this equation:-
$$\frac {(x-1)^2}{4}=1$$
we get,
$$(x-1)^2=4$$
Now try to do it.

17. Jul 7, 2011

### nae99

Re: logs

ok, but i thought (x-1) meant (x-1) (x-1)

18. Jul 7, 2011

### Saitama

Re: logs

I still didn't get you
$$(x-1)^2=(x-1)(x-1)=x^2-2x+1$$

19. Jul 7, 2011

### nae99

Re: logs

so i would end up with:

x^2-2x+1 = 4

20. Jul 7, 2011

### nae99

Re: logs

and then i would have to do the quadratic equation, right