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I was solving some problems and got to this logarithmic equation:

$$\log_2(\frac{36x^2-24x+4}{3x^2+8x+5})=0$$

How to solve this equation?

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  • $\begingroup$ How have you tried to solve the equation? $\endgroup$ – LucaMac Sep 9 '18 at 16:38
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Since $$\log_{2}(x) = \frac{\ln(x)}{\ln(2)}$$ then \begin{align} \log_2(\frac{36x^2-24x+4}{3x^2+8x+5}) &= 0 \\ \ln(\frac{36x^2-24x+4}{3x^2+8x+5}) &= 0 \\ \frac{36x^2-24x+4}{3x^2+8x+5} &= 1 \\ 36 x^2 - 24 x + 4 &= 3 x^2 + 8 x + 5 \\ 33 x^2 - 32 x -1 &= 0 \\ x^2 - 2\left(\frac{16}{33}\right) x - \frac{1}{33} &= 0 \\ x &= \frac{16}{33} \pm \sqrt{\frac{4 \cdot 16^2}{33^2} + \frac{4}{33}} \\ &= \frac{2}{33} \, (8 \pm \sqrt{16^2 + 33}) \\ &= \frac{2}{33} \, (8 \pm 17) \\ &= \left\{ \frac{50}{33}, - \frac{18}{33} \right\} \end{align}

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Hint: It must be $$\frac{36x^2-24x+4}{3x^2+8x+15}=1$$ Solve the equation $$33x^2-32x-11=0$$ One solution is given by $$x=\frac{1}{33} \left(16-\sqrt{619}\right)$$

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  • $\begingroup$ Aha thanks for the information. What kind of method would be best to use in the division? $\endgroup$ – Bili Debili Sep 9 '18 at 16:50
  • $\begingroup$ I would multiply by the denominator and solve a quadratic equation. $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '18 at 16:52

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