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.

Welcome back — been a while! Let's pick up where you left off. 3. Solve without using tables: (2x+5) + 3 = (x+11) Step 1: Use the logarithm property a + b = (ab) on the left side. (3(2x+5)) = (x+11) (6x+15) = (x+11) Step 2: Since the logarithms are equal, their arguments must be equal. 6x+15 = x+11 Step 3: Solve for x. 6x - x = 11 - 15 5x = -4 x = -(4)/(5) Step 4: Check the domain of the logarithms. The arguments must be positive. For (2x+5): 2x+5 > 0 2x > -5 x > -2.5. For (x+11): x+11 > 0 x > -11. Since x = -0.8, it satisfies both conditions (x > -2.5 and x > -11). The solution is valid. x = -(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 m^2, find its dimensions. Step 1: Write the formula for the area of a rectangle. A = length × width 77 = (x+4)(x-3) Step 2: Expand the right side of the equation. 77 = x^2 - 3x + 4x - 12 77 = x^2 + x - 12 Step 3: Rearrange the equation into a standard quadratic form ax^2+bx+c=0. x^2 + x - 12 - 77 = 0 x^2 + x - 89 = 0 Step 4: Solve the quadratic equation using the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. Here, a=1, b=1, c=-89. x = -1 ± sqrt(1^2 - 4(1)(-89))2(1) x = -1 ± sqrt(1 + 356)2 x = -1 ± sqrt(357)2 Step 5: Calculate the two possible values for x. sqrt(357) ≈ 18.89444 x_1 = (-1 + 18.89444)/(2) = (17.89444)/(2) ≈ 8.94722 x_2 = (-1 - 18.89444)/(2) = (-19.89444)/(2) ≈ -9.94722 Step 6: Determine the valid value for $x