## Logarithms

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 :)
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks

Blog Entries: 1
Recognitions:
Gold Member
Staff Emeritus
 Quote by Petkovsky 1. log((x-3)/(x+3)) <= 1
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.

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

Blog Entries: 1
Recognitions:
Gold Member
Staff Emeritus

## Logarithms

 Quote by thakid87 Am I seeing that right? Is that log [(x-3)/(x+3)] <= 1
 Quote by thakid87 And by <= you mean that it's less than one?
That generally means 'less than or equal to' (i.e. $\leq$).
 Quote by thakid87 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.
Hey, no need to apologise. Welcome to the Forums! Be sure to introduce yourself in the General Discussion forum .

 Quote by Hootenanny 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.
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?

Blog Entries: 1
Recognitions:
Gold Member
Staff Emeritus
 Quote by Petkovsky So we forget about the expansion and we have log[(x-3)/(x+3)] <= log10 Next (x-3)/(x+3) <= 10
Are you sure that that should be a 10?

Recognitions:
Gold Member
Staff Emeritus
 Quote by Petkovsky I have two problems if that's ok with you 1. log((x-3)/(x+3)) <= 1
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.

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

 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 :)

 Quote by Hootenanny Are you sure that that should be a 10?
Well if log(x) = 1
then 10$$^{1}$$ = x

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

 Quote by Hootenanny That's fairly easy to factor and so gives you two linear equations for a and b.
I cant see an easy way to factor it, honestly. Is it just me?
 Recognitions: Gold Member Homework Help Science Advisor 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 $6x^2-13x+6=0$?

Recognitions:
Gold Member
Staff Emeritus
 Quote by Petkovsky Well if log(x) = 1 then 10$$^{1}$$ = x From here x equals 10 I think it is ok...
So (x-3)/(x+3)= 10. Can you solve that for x?

Recognitions:
Gold Member
Staff Emeritus
 Quote by Petkovsky I cant see an easy way to factor it, honestly. Is it just me?
Yes, it really is!. It might help you to know that 3(3)+ 2(2)= 13.

Blog Entries: 1
Recognitions:
Gold Member
 Quote by Petkovsky Well if log(x) = 1 then 10$$^{1}$$ = x From here x equals 10 I think it is ok...