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I'm trying to solve this logarithmic equation for a while now, but I'm not getting any concrete solution.

$\log_{3x+8}(x^2 + 8x + 16) + \log_{x+4}(3x^2 + 20x + 32) = 7$

I defined the domain, converted to factors, applied the logarithmic rules, but I'm stil unable to solve it.

Domain: $x∈\left(-\frac{8}{3}, -\frac{7}{3}\right) ∪ \left(-\frac{7}{3}, +∞\right)$

Factoring: $\log_{3x+8}(x + 4)^2 + \log_{x+4}((3x + 8)(x + 4)) = 7$

Decomposition: $2 \log_{3x+8}(x + 4) + \log_{x+4}(3x + 8) + \log_{x+4}(x + 4) = 7$

After subtraction: $2 \log_{3x+8}(x + 4) + \log_{x+4}(3x + 8) = 6$

What now?

Any help would be appreciated! 🙏

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  • $\begingroup$ Welcome to math SE. Have a look at mathjax to typeset the equation $\endgroup$ Commented May 28 at 14:08
  • $\begingroup$ Welcome to math stack exchange. A couple of ways to improve your post (and likely get a better response): (1) Put your question directly into the post, using Mathjax. (2) Show specifically what you tried, rather than a general description of what you tried. $\endgroup$
    – paw88789
    Commented May 28 at 14:11
  • $\begingroup$ @AlainRemillard Thanks! Updated just now. $\endgroup$ Commented May 28 at 14:17
  • $\begingroup$ @paw88789 Sure! Thanks for the info $\endgroup$ Commented May 28 at 14:19

1 Answer 1

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Given: $$ \log_{3x+8}(x^2 + 8x + 16) + \log_{x+4}(3x^2 + 20x + 32) = 7 $$

$$ \log_{3x+8}((x + 4)^2) + \log_{x+4}((3x + 8)(x + 4)) = 7 $$

$$ 2 \log_{3x+8}(x + 4) + \log_{x+4}(3x + 8) + \log_{x+4}(x + 4) = 7 $$

$$ 2 \log_{3x+8}(x + 4) + \log_{x+4}(3x + 8) = 6 $$

Let $ \log_{3x+8}(x + 4) = a $ and by change of base formula $ \log_{x+4}(3x + 8) = \frac{1}{a} $

The equation becomes $$ 2a + \frac{1}{a} = 6 $$ $$ 2a^2 - 6a + 1 = 0 $$

$$ a = \frac{3 \pm \sqrt{7}}{2} $$

$$ \log_{3x+8}(x + 4) = \frac{3 + \sqrt{7}}{2} \quad \text{or} \quad \log_{3x+8}(x + 4) = \frac{3 - \sqrt{7}}{2} $$

Solving numerically you get $x\approx -2.26117 $

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  • $\begingroup$ Cool. Thank you so much. How did you solve this numerically tho? $\endgroup$ Commented May 28 at 14:50
  • $\begingroup$ enchanted silicon runes $\endgroup$
    – vallev
    Commented May 29 at 18:20
  • $\begingroup$ Bahahaha, makes sense $\endgroup$ Commented May 29 at 19:30

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