Solve the following equations: (a) log(2x+5) + log3 = log(x+11); (b) The length of a rectangular hall is (x+4) m and the width is (x-3) m. If its area is 77 m2, find its dimensions.

$$\log(3(2x+5)) = \log(x+11)$$ $$\log(6x+15) = \log(x+11)$$

$$6x - x = 11 - 15$$ $$5x = -4$$ $$x = -\frac{4}{5}$$

The arguments must be positive.
For $\log(2x+5)$: $2x+5 > 0 \Rightarrow 2x > -5 \Rightarrow x > -2.5$.
For $\log(x+11)$: $x+11 > 0 \Rightarrow x > -11$.
Since $x = -0.8$, it satisfies both conditions ($x > -2.5$ and $x > -11$).
The solution is valid.

$$\boxed{x = -\frac{4}{5}}$$

4. The length of a rectangular hall is $(x+4)$ m and the width is $(x-3)$ m. If its area is $77 \text{ m}^2$, find its dimensions.

$$A = \text{length} \times \text{width}$$ $$77 = (x+4)(x-3)$$

$$77 = x^2 - 3x + 4x - 12$$ $$77 = x^2 + x - 12$$

$$x^2 + x - 12 - 77 = 0$$ $$x^2 + x - 89 = 0$$

Here, $a=1$, $b=1$, $c=-89$.

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-89)}}{2(1)}$$ $$x = \frac{-1 \pm \sqrt{1 + 356}}{2}$$ $$x = \frac{-1 \pm \sqrt{357}}{2}$$

$$\sqrt{357} \approx 18.89444$$ $$x_1 = \frac{-1 + 18.89444}{2} = \frac{17.89444}{2} \approx 8.94722$$ $$x_2 = \frac{-1 - 18.89444}{2} = \frac{-19.89444}{2} \approx -9.94722$$