Division of polynomials in logarithm

Question

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?

Answer 1

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}

Answer 2

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 $$x=\frac{1}{33} \left(16-\sqrt{619}\right)$$