3
$\begingroup$

The equation that I'm trying to solve is: $$\log _{5x+9}(x^2+6x+9)+\log _{x+3}(5x^2+24x+27)=4$$

Using algebra and principles of logarithms I managed to get the equation down to $$\frac{2\left(\log \left(x+3\right)\right)^2+\left(\log \left(5x+9\right)\right)^2}{\log \left(x+3\right)\log \left(5x+9\right)}=3$$ assuming that everything I've done so far is correct. I'm not sure what to do from here. I think there might be some kind of perfect square to factor but I'm not sure how I would even factor it from this situation (or if it would help).

Any suggestions?

$\endgroup$
1
  • $\begingroup$ Thanks for all the help and not just giving away the answer. $\endgroup$
    – Jonathan
    Aug 25, 2015 at 18:57

3 Answers 3

5
$\begingroup$

For the sake of simplification, set $a=\log (x+3)$ and $b=\log (5x+9)$ then the equation you obtained writes as $2a^2-3ab+b^2=0$. Using standard factoring this is $(a-b)(2a-b)=0$. I think you can do the rest!

$\endgroup$
1
  • $\begingroup$ Okay, thanks I think I can solve it now. $\endgroup$
    – Jonathan
    Aug 25, 2015 at 18:40
0
$\begingroup$

Hint:

$$ 5x^2+24x+27=(x+3)(5x+9) $$ so your equation becomes: $$ 2\log_{5x+9}(x+3)+\log_{x+3}[(x+3)(5x+9)]=4 $$

now use $\log_a a=1$ and $\log_a b=\dfrac{1}{\log_b a}$

$\endgroup$
0
$\begingroup$

Note that $5x^2+24x+9=(x+3)(5x+9)$, then the equation is equivalent to $2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$. Let $\log_{5x+9}(x+3)=z$. So the equation is equivalent to $2z^2+1/z-3=0$ and thus $(z-1)(2z-1)=0$. So you can conclude.

$\endgroup$
1
  • $\begingroup$ This is a good solution. It is much shorter than the one I used. $\endgroup$
    – Jonathan
    Aug 25, 2015 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.