Solving $\log_6(2x-3)+\log_6(x+5)=\log_3x$ Solve for $x$
I have an equation that I have been working on solving; I know the solution, but I cannot get to it myself. Almost every simplification I do reverts back to a previous step. Can anyone show me how to solve for $x$ in this equation?
Equation:
$$\log_6(2x-3)+\log_6(x+5)=\log_3x$$
Solution:
$$x ≅ \frac{3347}{2000} ≅ 1.6735$$

Note: upon further analysis of the answer, while close, it does not seem to be the exact solution.

What I Have Tried So Far
$$\log_6(2x-3) + \log_6(x + 5) = \log_3x$$
$$\frac{\log(2x-3)}{\log6} + \frac{\log(x + 5)}{\log6} = \frac{\log x}{\log3}$$
$$\log3 \cdot \log(2x-3) + \log3 \cdot \log(x + 5) = \log6 \cdot \log x$$
$$\log3 \cdot \log \left[(2x - 3)(x + 5)\right] = \log6 \cdot \log x$$
$$\frac{\log \left[(2x - 3)(x + 5)\right]}{\log_3 10} = \frac{\log6}{\log_x10}$$
$$\log_x10 \cdot \log \left[(2x - 3)(x + 5)\right] = \log_3 10 \cdot \log6$$
$$\log_x \left[(2x - 3)(x + 5)\right] = \log_3 6$$
$$\log_x3 \cdot \log_x \left[(2x - 3)(x + 5)\right] = \frac{\log_3 6}{\log_3 x}$$
$$\log_x \left[(2x - 3)(x + 5)\right]^{\ \log_x3} = \log_x 6$$
$$\left[(2x - 3)(x + 5)\right]^{\ \log_x3} = 6$$
$$(2x - 3)(x + 5) = x^{\log_3 6}$$

I know these steps aren't really working towards the solution at points; I was sort of just playing around with the equation. Regardless, I really don't know how to go about moving forward from here.


Taking the base-$6$ antilogarithm,
$$(2x-3)(x+5)=6^{\log_3x}=e^{\ln6\ln x/\ln3}=x^{\ln6/\ln3}.$$
Because of the irrational exponent, there is no closed-form solution and you need to use a numerical method.
