Solve the logarithmic equation: x + (2x − 5) = 3

$$2x^2 - 5x = 3$$ $$2x^2 - 5x - 3 = 0$$

We can use factoring or the quadratic formula. Let's factor. We need two numbers that multiply to $2 \times (-3) = -6$ and add to $-5$. These numbers are $-6$ and 1.

$$2x^2 - 6x + x - 3 = 0$$ $$2x(x - 3) + 1(x - 3) = 0$$ $$(2x + 1)(x - 3) = 0$$

This gives two possible solutions for $x$:

$$2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ $$x - 3 = 0 \implies x = 3$$

The argument of a logarithm must be positive.
For $x = -\frac{1}{2}$:
The term $\log x$ becomes $\log \left(-\frac{1}{2}\right)$, which is undefined for real numbers. So, $x = -\frac{1}{2}$ is not a valid solution.

For $x = 3$:
The term $\log x$ becomes $\log 3$, which is valid.
The term $\log(2x - 5)$ becomes $\log(2(3) - 5) = \log(6 - 5) = \log 1$, which is valid.
Therefore, $x = 3$ is the only valid solution.

The value of $x$ is $\boxed{3}$.