Any idea how to solve this equation?

Any idea how to solve this equation? $$x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3)=3\log_{3}\frac{x}{2x+3}$$

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It is not homework. I am professor. –  Chun-Yue Dec 1 '11 at 13:28
I am sorry. I did mean no offense. I sould have taken a closer look at the problem. From a first glimpse it seemed like converting the different parts to $\log_e$ and spliting the logarithms, everything should cancel nicely. That's why I suspected it was homework in the first place. –  fabee Dec 1 '11 at 14:24
A professor that doesn't seem to reward the work of his good students. –  Raskolnikov Dec 1 '11 at 14:45
@Chun-Yue As Raskolnikov is hinting above, and as I have mentioned before, please accept an answer which you find acceptable. You have asked 9 questions, and have not accepted any answers. This site functions much better when you accept answers to the questions you ask. You can go back to any of your previous questions and click the check mark to the left of the answer you would like to accept. –  process91 Dec 1 '11 at 15:17

Simplifying this equation by the step below: $(2x^2+3)\log_{9}(2x+3)=(2x^2+3)\frac{\log_{3}(2x+3)}{\log_{3}9}=\frac{2x^2+3}{2}\log_{3}(2x+3)$

Now the equation is : $x^2\log_{3}x^2-\frac{2x^2+3}{2}\log_{3}(2x+3)=3\log_{3}\frac{x}{2x+3}$

Eliminate the log symbol, we get: $\frac{x^{2x^2}}{(2x+3)^{\frac{2x+3}{2}}}=(\frac{x}{2x+3})^3$

Reduce and simplify: $x^{2x^2-3}=(2x+3)^{\frac{2x^2-3}{2}}$

Now, we see $2x^2-3=0$ or $x=(2x+3)^{\frac{1}{2}}$, namely, $x^2-2x-3=0$

So the roots are $x_1=\sqrt{\frac{3}{2}}, x_2=-\sqrt{\frac{3}{2}},x_3=-1,x_4=3$

Note that log function requires positive values, so $x_2$ and $x_3$should be discarded by substituting them into the original equation.

Conclusion: the roots are $x_1=\sqrt{\frac{3}{2}}, x_2=3$

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Note that $\log_9(x) = \log_3(x)/\log_3(9) = \frac{1}{2}\log_3(x)$.

We need $2x+3\gt 0$, and hence $x\gt 0$, so that all logarithms can be evaluated.

So we can simplify the equation using the basic properties of the logarithms

\begin{align*} x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3) &= 3\log_{3}\left(\frac{x}{2x+3}\right)\\ 2x^2\log_3(x) - \frac{2x^2+3}{2}\log_3(2x+3) &= 3\log_3(x) - 3\log_3(2x+3)\\ (2x^2-3)\log_3(x) &= \left(\frac{2x^2+3}{2} - 3\right)\log_3(2x+3)\\ (2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3). \end{align*} From this, it is clear that any solutions to $2x^2-3=0$ will satisfy the equation. The solutions are $x=\sqrt{3/2}$ and $x=-\sqrt{3/2}$; but the latter cannot be used in the original equation. So one possibility is $$x = \sqrt{\frac{3}{2}}.$$

If $2x^2-3\neq 0$, then cancelling we get: \begin{align*} (2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3)\\ \log_3(x) &= \frac{1}{2}\log_3(2x+3)\\ 2\log_3(x) - \log_3(2x+3) &= 0\\ \log_3\left(\frac{x^2}{2x+3}\right) &= 0\\ \frac{x^2}{2x+3} &= 1\\ x^2 -2x - 3 &= 0\\ (x-3)(x+1) &=0. \end{align*} Again, $x=-1$ cannot be used. So the only other solution is $x=3$.

So the solutions are $x=3$ and $x=\sqrt{\frac{3}{2}}$.

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