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.

• What are the steps you have tried? Commented Mar 13, 2019 at 0:22
• Do you have reason to believe you will get a nice closed form for the answer? Mixing logs with different bases generally doesn't end well. Commented Mar 13, 2019 at 0:22
• Are you sure about the $\log_3$ ?
– user65203
Commented Mar 13, 2019 at 0:23
• Mind to tell us the solution that you have ?
– user65203
Commented Mar 13, 2019 at 0:26
• I also have updated the question with the steps I have already taken to try to solve the equation @MarsPlastic. Commented Mar 13, 2019 at 0:52

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.

• Why this downvote ?
– user65203
Commented Mar 13, 2019 at 0:33

Just for the fun of it !

Since heropup already gave the answer, let us do the same using one single iteration using high order methods with $$x_0 = \frac{3347}{2000}$$

$$\left( \begin{array}{ccc} n & x_1 & \text{Method} \\ 1 & \color{blue}{1.67351617}525914512770936502001715480227532437 & \text{Newton} \\ 2 & \color{blue}{1.6735161761242}2229623976832154151388262101881 & \text{Halley}\\ 3 & \color{blue}{1.67351617612426023}725117282565297065627253993 & \text{Householder}\\ 4 & \color{blue}{1.6735161761242602388483}2362357774862967541409 & \text{no name}\\ 5 & \color{blue}{1.6735161761242602388483916}1928419708807692011 & \text{no name}\\ 6 & \color{blue}{1.673516176124260238848391622220}46176652833341 & \text{no name}\\ 7 & \color{blue}{1.67351617612426023884839162222058963}357352924 & \text{no name}\\ 8 & \color{blue}{1.673516176124260238848391622220589639170}05729 & \text{no name} \end{array} \right)$$