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

The logarithm property $\log a - \log b = \log(a/b)$ can be applied to both sides of the equation.

$$\log\left(\frac{2x-11}{2}\right) = \log\left(\frac{3}{x}\right)$$

If $\log A = \log B$, then $A = B$.

$$\frac{2x-11}{2} = \frac{3}{x}$$

Multiply both sides by $2x$ to eliminate the denominators.

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

Rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$.

$$2x^2 - 11x - 6 = 0$$

We look for two numbers that multiply to $2 \times -6 = -12$ and add to $-11$. These numbers are $-12$ and 1.

$$2x^2 - 12x + x - 6 = 0$$

Factor by grouping:

$$2x(x-6) + 1(x-6) = 0$$ $$(2x+1)(x-6) = 0$$

This gives two potential solutions for $x$:

$$2x+1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ $$x-6 = 0 \implies x = 6$$

The argument of a logarithm must be positive. From the original equation, we have two conditions:

1. $2x-11 > 0 \implies 2x > 11 \implies x > 5.5$
2. $x > 0$

Let's check our potential solutions:

• For $x = -\frac{1}{2}$: This value does not satisfy $x > 5.5$ or $x > 0$. Therefore, $x = -\frac{1}{2}$ is an extraneous solution.
• For $x = 6$: This value satisfies both $x > 5.5$ (since $6 > 5.5$) and $x > 0$. Therefore, $x=6$ is the valid solution.