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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?

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

3 Answers 3

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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!

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  • $\begingroup$ Okay, thanks I think I can solve it now. $\endgroup$
    – Jonathan
    Aug 25, 2015 at 18:40
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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 $\log_a b=\dfrac{1}{\log_b a}$

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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.

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

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