log(x+2) + log(2x−1) = log 15

$$\log (x+2) + \log (2x-1) = \log ((x+2)(2x-1))$$

So the equation becomes:

$$\log ((x+2)(2x-1)) = \log 15$$

$$2x^2 - x + 4x - 2 = 15$$ $$2x^2 + 3x - 2 = 15$$

Subtract 15 from both sides:

$$2x^2 + 3x - 17 = 0$$

Here, $a=2$, $b=3$, $c=-17$.

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-17)}}{2(2)}$$ $$x = \frac{-3 \pm \sqrt{9 + 136}}{4}$$ $$x = \frac{-3 \pm \sqrt{145}}{4}$$

This gives two potential solutions:

$$x_1 = \frac{-3 + \sqrt{145}}{4} \quad \text{and} \quad x_2 = \frac{-3 - \sqrt{145}}{4}$$

The arguments of the original logarithms must be positive, i.e., $x+2 > 0$ and $2x-1 > 0$.
This means $x > -2$ and $x > \frac{1}{2}$. So, we must have $x > \frac{1}{2}$.
Let's approximate $\sqrt{145}$. Since $12^2 = 144$, $\sqrt{145}$ is slightly greater than 12, approximately 12.04.

For $x_1$:

$$x_1 = \frac{-3 + \sqrt{145}}{4} \approx \frac{-3 + 12.04}{4} = \frac{9.04}{4} = 2.26$$

This value is greater than $\frac{1}{2}$, so it is a valid solution.

For $x_2$:

$$x_2 = \frac{-3 - \sqrt{145}}{4} \approx \frac{-3 - 12.04}{4} = \frac{-15.04}{4} = -3.76$$

This value is not greater than $\frac{1}{2}$ (it's less than -2 and less than 1/2), so it is an extraneous solution and must be rejected.

The only valid solution is:

$$x = \frac{-3 + \sqrt{145}}{4}$$