Find the value of x that satisfies the equation Log(2x−11)−log2=log3−log x

$$x(2x-11) = 2(3)$$ $$2x^2 - 11x = 6$$

Here, $a=2$, $b=-11$, and $c=-6$.

$$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(2)(-6)}}{2(2)}$$ $$x = \frac{11 \pm \sqrt{121 + 48}}{4}$$ $$x = \frac{11 \pm \sqrt{169}}{4}$$ $$x = \frac{11 \pm 13}{4}$$

$$x_1 = \frac{11 + 13}{4} = \frac{24}{4} = 6$$ $$x_2 = \frac{11 - 13}{4} = \frac{-2}{4} = -\frac{1}{2}$$

The argument of a logarithm must be positive.
For $x=6$:
$2x-11 = 2(6)-11 = 12-11 = 1 > 0$
$x = 6 > 0$
This solution is valid.

For $x=-\frac{1}{2}$:
$2x-11 = 2(-\frac{1}{2})-11 = -1-11 = -12 < 0$
The argument of the logarithm cannot be negative. This solution is extraneous.

The only valid solution is $x=6$.

$$6$$