What are some strategies for solving logarithmic equations?

[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 don't know what to do from here. any help would be apreciated. thanks :)

 

[Hootenanny]

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.

 

[thakid87]

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.

 

[Hootenanny]

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

That's how I read it.

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

That generally means 'less than or equal to' (i.e. $\leq$).

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 .

 

[Petkovsky]

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?

 

[Hootenanny]

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?

 

[HallsofIvy]

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= 3^1/x and b= 2^1/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 don't know what to do from here. any help would be apreciated. thanks :)

 

[Petkovsky]

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...

 

[Petkovsky]

That's fairly easy to factor and so gives you two linear equations for a and b.

I can't see an easy way to factor it, honestly. Is it just me?

 

[quantumdude]

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 $6x^2-13x+6=0$?

 

[HallsofIvy]

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?

 

[HallsofIvy]

I can't 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.

 

[Hootenanny]

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

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

Sorry, my bad, I misread the OP.