1. Jun 23, 2008

### Petkovsky

I have two problems if that's ok with you

1. log((x-3)/(x+3)) <= 1

2. 6*9^(1/x) - 13*6^(1/x) + 6*4^(1/x) = 0

So for the first one i didnt get too far. I divided the logarithm and i got:
log(x-3) + log(x+3) <= 1

For the second one i used logarithms and received:
log6 + 1/x*log9 - log13 - 1/x*log6 + log6 + 1/x*log4 = 0
Next:
1/x(log9-log6+log4) + log6 - log13 + log6 = 0
Then:
1/x*(log6) + log(36/13) = 0

and i dont know what to do from here. any help would be apreciated. thanks :)

2. Jun 23, 2008

### Hootenanny

Staff Emeritus
Let's start with the first one. Start by finding the value of y such that log(y)=1.

You also need to be careful with your logarithm laws, your expansion is incorrect.

Last edited: Jun 23, 2008
3. Jun 23, 2008

### thakid87

Am I seeing that right? Is that log [(x-3)/(x+3)] <= 1

And by <= you mean that it's less than one?

Sorry guys. Getting used the forums here. I also thought that a good way to brush up my "skills" is to try and help other people.

Thanks.

4. Jun 23, 2008

### Hootenanny

Staff Emeritus
That generally means 'less than or equal to' (i.e. $\leq$).
Hey, no need to apologise. Welcome to the Forums! Be sure to introduce yourself in the General Discussion forum .

5. Jun 23, 2008

### Petkovsky

Sorry it's just a typo. The sign in between is supposed to be a minus.

Btw u helped me a lot with that hint so let me just chech if i got it.

So we forget about the expansion and we have log[(x-3)/(x+3)] <= log10
Next
(x-3)/(x+3) <= 10
x-3 <= 10x + 30

x<= -33/9 << is this ok?

6. Jun 23, 2008

### Hootenanny

Staff Emeritus
Are you sure that that should be a 10?

7. Jun 23, 2008

### HallsofIvy

Staff Emeritus
As Hootenanny said, first solve log(y)= 1, then use y= (x-3)/(x+3). Remember than an inequality can change from "<" to ">" only at points where both sides are equal or where the function is not defined.

\
9= 32, 6= 3(2) and 4= 22. If you let a= 31/x and b= 21/x, this is 6a2- 13ab+ 6b2= 0. That's fairly easy to factor and so gives you two linear equations for a and b.

8. Jun 23, 2008

### Petkovsky

Well if log(x) = 1
then 10$$^{1}$$ = x

From here x equals 10
I think it is ok...

9. Jun 23, 2008

### Petkovsky

I cant see an easy way to factor it, honestly. Is it just me?

10. Jun 23, 2008

### Tom Mattson

Staff Emeritus
It is easy to factor, but if you don't see it you can always fall back on the quadratic formula. What are the solutions of $6x^2-13x+6=0$?

11. Jun 24, 2008

### HallsofIvy

Staff Emeritus
Re: Logarithms

So (x-3)/(x+3)= 10. Can you solve that for x?

12. Jun 24, 2008

### HallsofIvy

Staff Emeritus
Re: Logarithms

Yes, it really is!. It might help you to know that 3(3)+ 2(2)= 13.

13. Jun 24, 2008

### Hootenanny

Staff Emeritus
Re: Logarithms