1. **State the problem:** Solve the equation $$\log(2x + 3) + \log(x + 2) - 1 = 0$$ for $x$. 2. **Use the logarithm property:** The sum of logarithms is the logarithm of the product: $$\log a + \log b = \log(ab)$$ So, $$\log(2x + 3) + \log(x + 2) = \log\big((2x + 3)(x + 2)\big)$$ 3. **Rewrite the equation:** $$\log\big((2x + 3)(x + 2)\big) - 1 = 0$$ 4. **Isolate the logarithm:** $$\log\big((2x + 3)(x + 2)\big) = 1$$ 5. **Convert from logarithmic to exponential form:** Since the base of the logarithm is 10 (common log), $$ (2x + 3)(x + 2) = 10^1 = 10 $$ 6. **Expand the product:** $$ 2x^2 + 4x + 3x + 6 = 10 $$ $$ 2x^2 + 7x + 6 = 10 $$ 7. **Bring all terms to one side:** $$ 2x^2 + 7x + 6 - 10 = 0 $$ $$ 2x^2 + 7x - 4 = 0 $$ 8. **Solve the quadratic equation using the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=2$, $b=7$, $c=-4$. Calculate the discriminant: $$ \Delta = 7^2 - 4 \times 2 \times (-4) = 49 + 32 = 81 $$ Calculate the roots: $$ x = \frac{-7 \pm \sqrt{81}}{2 \times 2} = \frac{-7 \pm 9}{4} $$ 9. **Find the two solutions:** $$ x_1 = \frac{-7 + 9}{4} = \frac{2}{4} = 0.5 $$ $$ x_2 = \frac{-7 - 9}{4} = \frac{-16}{4} = -4 $$ 10. **Check the domain restrictions:** The arguments of the logarithms must be positive: $$ 2x + 3 > 0 \Rightarrow x > -\frac{3}{2} = -1.5 $$ $$ x + 2 > 0 \Rightarrow x > -2 $$ So the domain is $x > -1.5$. Check $x_1 = 0.5$: valid. Check $x_2 = -4$: invalid (less than -1.5). 11. **Final answer:** $$ \boxed{0.5} $$