Solving $\log_6(2x-3)+\log_6(x+5)=\log_3x$

If $x \in \mathbb R$, the equation $$\log_6 (2x-3) + \log_6 (x+5) = \log_3 x$$ requires $x > 3/2$. Under such an assumption, the LHS becomes $$\log_6 (2x-3)(x+5),$$ and the RHS, using the change-of-base formula, is $$\log_3 x = \frac{\log_6 x}{\log_6 3}.$$ Thus $$(2x-3)(x+5) = 6^{\log_6 x / \log_6 3} = x^{\log_3 6} = x^{\log_3 3 + \log_3 2} = x^{1 + \log_3 2}.$$

Let $a = \log_3 2 < 1$. Then to seek a numerical root, we apply Newton's method to $$f(x) = 2x^2 - x^{a+1} + 7x - 15$$ by computing iterates of $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{2x_n^2 - x_n^{a+1} + 7x_n - 15}{4x_n - (a+1)x_n^a + 7}.$$ A suitable initial guess has already been provided, namely $x_0 = \frac{3347}{2000}$, from which we iterate with a computer to obtain to 75 digits of precision $$x_0 = \color{green}{1.6735}0000000000000000000000000000000000000000000000000000000000000000000000 \\ x_1 = \color{green}{1.6735161761}6028420105594976526962483014865547809090912454018791698117840074 \\ x_2 = \color{green}{1.673516176124260238848}57028078132556340450344946364614508901732583022202367 \\ x_3 = \color{green}{1.67351617612426023884839162222058963917030308}792976477026920554313588396492 \\ x_4 = \color{green}{1.67351617612426023884839162222058963917030308353547657451861889646281230396} \\ x_5 = \color{green}{1.67351617612426023884839162222058963917030308353547657451861889646281230396} $$ where green digits indicate correct values, demonstrating the rapid convergence of the iterates. Moreover, we are assured that this is the unique root, since $f$ is a monotonically increasing function on $x > 3/2$ (its derivative being trivially greater than $0$ on this interval).

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.