Solve the Logarithmic Equation: (2log₃2 + log₃x)log₂3 - log₂x = 3x-7

$(2\log_32+\log_3x)\log_23-\log_2x=3x-7$

$x=\text{?}$

Let's solve the given equation step by step:

We start with:

$(2\log_3 2 + \log_3 x)\log_2 3 - \log_2 x = 3x - 7$

Firstly, use the change of base formula to convert $\log_2 3$ to base 3:

$\log_2 3 = \frac{\log_3 3}{\log_3 2} = \frac{1}{\log_3 2}$

Substitute this expression into the original equation:

$(2\log_3 2 + \log_3 x)\left(\frac{1}{\log_3 2}\right) - \log_2 x = 3x - 7$

Simplify the first term:

$\frac{2\log_3 2 + \log_3 x}{\log_3 2} = 2 + \frac{\log_3 x}{\log_3 2}$

Thus, the equation becomes:

$2 + \frac{\log_3 x}{\log_3 2} - \log_2 x = 3x - 7$

Convert $\log_2 x$ to base 3 using change of base:

$\log_2 x = \frac{\log_3 x}{\log_3 2}$

Substitute back into the equation:

$2 + \frac{\log_3 x}{\log_3 2} - \frac{\log_3 x}{\log_3 2} = 3x - 7$

The middle terms cancel out, simplifying to:

2 = 3x - 7

Solving for $x$:

Add 7 to both sides:

$9 = 3x$

Divide by 3:

$x = 3$

Thus, the solution to the problem is $x = 3$.