solve

[tomas]

[Phi for All]

Listen, tomas, you're going to need to give us a bit more background on this. Is this for homework? We don't do your homework for you. We can help aim you in the right direction but you need to talk to us here.

 

What's going on?

[tomas]

these is from another forum and i dont no how i can solve it

 

thanks

[Phi for All]

OK, thanks for the reply. Is it for school homework?

[tomas]

no these is so defeclt , i thenk it is not school homework

[Riogho]

I've been applying my n00b skills to this problem and I'm stumped. I'd be curious as to the answer as well. More importantly however, how you got it. I just keep getting myself caught in circles.

[Kyrisch]

Well [math] 9 = 3^2 [/math] and your log bases are 3... There's a clue .

 

Also, I'm having trouble putting in the log base subscript with LaTeX... how do you do that?

Edited June 7, 2008 by Kyrisch

[Bignose]

Well, there aren't any restrictions on how to solve it. So I graphed it. There is an answer. I don't know if there are any pure analytical ways to get to it -- but numerical solutions are just as valid as any other method.

[Kyrisch]

[math] \log_3{(4x^2-9)}^{\log_3{(x+6)}}=3^2[/math]

 

[math] 3^{3^2} = (4x^2 - 9)^{\log_3{(x+6)}} [/math]

 

[math] \log_{4x^2-9}{3^9} = \log_3{(x+6)} [/math]

 

[math] 9\log_{4x^2-9}{3} = \log_3{(x+6)} [/math]

 

[math] 9 \frac{\log_3{3}}{\log_3{(4x^2-9)}} = \log_3{(x+6)} [/math]

 

[math] \frac{9}{\log_3{(4x^2-9)}} = \log_3{(x+6)} [/math]

 

...is as far as I can get

[Riogho]

I got it all on one line, but my skills are really rusty.

 

Forgive me, I don't know LateX either.

 

log(x+6) log(2log(4x-9)) = log(9)log(3)log(3)

 

That's as far as I got