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This was my logic: $$ \log_{3}(2x+5) = [\log_{9}(4x+1)]^2 \iff \log_{3}(2x+5) = \left(\dfrac{\log_{3}(4x+1)}{\log_{3}9}\right)^2 \iff \log_{3}(2x+5) = \dfrac{\log_{3}(4x+1) \cdot \log_{3}(4x+1)}{4} \iff 4 \cdot \log_{3}(2x+5) - \log_{3}(4x+1) \cdot \log_{3}(4x+1) = 0 $$

But now factorization isn't helping me. I don't see if there is a way to join the logs or if there is a convenient substitution. Probably there is a typo at the source of the equation, but I got curious and wanted to solve this.

ps: the entire log is squared, and it turns hard to solve for that reason.

Carlos Oliveira
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    Well by convention $\log_9 (4x+1)^2$ is not $[\log_9(4x+1)]^2$, but it should be $\log_9 [(4x+1)^2] = 2\log_9(4x+1)$. – Hw Chu Mar 16 '18 at 00:42
  • @HwCHu: Wolfram seems to not use that convention - it interprets $\log(x)^2$ as $(\log(x))^2$. The OP's math seems to be in agreement with Wolfram's convention. Probably best to ask him for clarification as to whether $\log_9 (4x+1)^2 = \log_9( (4x+1)^2)$ or $\log_9 (4x+1)^2 = (\log_9(4x+1))^2$. – Jason DeVito - on hiatus Mar 16 '18 at 01:07
  • @JasonDeVito It's not specified, so it's assumed that you are raising the power to the argument of the logarithm. –  Mar 16 '18 at 01:11
  • @idk: That convention is unfamiliar to me. For example, if I saw $\sin(x)^2$, I would think $\sin^2(x)$, not $\sin(x^2)$. This seems to be the same reasoning with $\log$. That said, $\log$ is not a function I work with on a regular basis, so I could have just missed all conventions about it. – Jason DeVito - on hiatus Mar 16 '18 at 01:13
  • @JasonDeVito Then it would depend, and the original post has to specify which one. –  Mar 16 '18 at 01:15
  • Hi, the power applies to the entire log. I have added brackets to clarify that. I know it becomes very easy if we square only the input of the log function. Thank you for your attention. – Carlos Oliveira Mar 16 '18 at 16:20
  • I am starting to think that math doesn't provide what is necessary for my case... It is a little shame, but we will overcome it. – Carlos Oliveira Mar 18 '18 at 23:47
  • We will definitely find a solution. :) – dssknj Mar 19 '18 at 17:26

5 Answers5

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I might be wrong, but $$\log_3 (2x+5)=\log_9 (4x+1)^2 = 2\log_9 (4x+1)$$ implies $$2x+5=3^{2\log_9 (4x+1)}=(3^2)^{\log_9 (4x+1)}=9^{\log_9 (4x+1)}=4x+1.$$

Rócherz
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Note that $\log_a (b)^2=2\log_a (b)=2\cdot\dfrac{\log b}{\log a}\ne\left(\dfrac{\log b}{\log a}\right)^2$

Apply this to your second step, which I believe you performed incorrectly.

If you do this correctly, you will eventually reach that:

$\log_3(2x+5)=2\cdot\dfrac{\log_3(4x+1)}{\log_3 9}$

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Alternatively: $$\log_{3} (2x+5)=\log_{3^2} (4x+1)^2=\frac{2}{2} \log_{3} |4x+1| \Rightarrow \\ \pm(2x+5)=4x+1 \Rightarrow \\ x=2;-1.$$

farruhota
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After search a while, it seems like only numerical methods can provide the answer to my question. Although I was wanting some symbolic manipulation, that seems the way for now.

Carlos Oliveira
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Since you loog for the solution of $$\log_{3}(2x+5) = \left(\log_{9}(4x+1)\right)^2$$ (let me switch to natural logarithms) consider that you look for the zero of function $$f(x)=\frac{\log (2 x+5)}{\log (3)}-\frac{\log ^2(4 x+1)}{\log ^2(9)}$$ $$f'(x)=\frac{2}{(2 x+5) \log (3)}-\frac{8 \log (4 x+1)}{(4 x+1) \log ^2(9)}$$ Use inspection and after a "few" attempts, you will see that $$f(10)=\frac{\log (25)}{\log (3)}-\frac{\log ^2(41)}{\log ^2(9)}\approx +0.0734432$$ $$f(11)=\frac{\log (27)}{\log (3)}-\frac{\log ^2(45)}{\log ^2(9)}\approx -0.0015104$$

So, the solution is very close to $x_0=11$. Since there is no analytical solution, use Newton method and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 11.00000000 \\ 1 & 10.97923902 \\ 2 & 10.97925143 \end{array} \right)$$ If you perform only one ieration of Newton method, you would get $$x_1=11-\frac{135} 2\frac{ \log ^2(9) \log (27)-\log (3) \log ^2(45)}{ 5 \log ^2(9)-12 \log (3) \log (45)}$$ that is to say $$x_1=11-\frac{135} 8 \frac{8 \log ^2(3)-\log ^2(5)-4 \log (3) \log (5)}{\log (375)}$$

Claude Leibovici
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