# Unable to solve a logarithmic equation

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! 🙏

• Welcome to math SE. Have a look at mathjax to typeset the equation Commented May 28 at 14:08
• 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. Commented May 28 at 14:11
• @AlainRemillard Thanks! Updated just now. Commented May 28 at 14:17
• @paw88789 Sure! Thanks for the info Commented May 28 at 14:19

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 $$x\approx -2.26117$$

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