Q4. (i) Find the exact solution of the equation \[ 8^{2 x+1}=6 \] giving your answer in the form \( a+b \log _{2} 3 \), where \( a \) and \( b \) are constants to be found. (ii) Using the laws of logarithms, solve \[ \log _{5}(7-2 y)=2 \log _{5}(y+1)-1 \]

We solve each part step by step. ───────────────────────────── Part (i) We need to solve   8^(2x + 1) = 6 Step 1. Express 8 as a power of 2. Since 8 = 2³, we can rewrite the left side:   (2³)^(2x + 1) = 2^(3(2x + 1)) = 2^(6x + 3). Thus the equation becomes   2^(6x + 3) = 6. Step 2. Take logarithms with base 2 of both sides:   6x + 3 = log₂6. Step 3. Solve for x:   6x = log₂6 – 3   x = (log₂6 – 3) / 6. Step 4. Write log₂6 in a form that involves log₂3. Note that 6 = 2 × 3, so   log₂6 = log₂2 + log₂3 = 1 + log₂3. Substitute back:   x = (1 + log₂3 – 3) / 6 = (log₂3 – 2) / 6. This can be rearranged to emphasize the required form a + b·log₂3:   x = –2/6 + (1/6)·log₂3 = –1/3 + (1/6)·log₂3. Thus, the constants are a = –1/3 and b = 1/6. ───────────────────────────── Part (ii) We need to solve   log₅(7 – 2y) = 2·log₅(y + 1) – 1. Step 1. Use the logarithm laws. Express the right-hand side as a single logarithm. Notice:   2·log₅(y + 1) = log₅((y + 1)²). Also, recall that subtracting 1 is equivalent to subtracting log₅5 because log₅5 = 1. Hence,   2·log₅(y + 1) – 1 = log₅((y + 1)²) – log₅5 = log₅[(y + 1)²/5]. So the equation becomes   log₅(7 – 2y) = log₅[(y + 1)²/5]. Step 2. Equate the arguments (the expressions inside the logarithms) provided they are positive:   7 – 2y = (y + 1)² / 5. Step 3. Multiply both sides by 5:   5(7 – 2y) = (y + 1)²   35 – 10y = y² + 2y + 1. Step 4. Rearranging to set the quadratic to zero:   0 = y² + 2y + 1 + 10y – 35   0 = y² + 12y – 34. This is the quadratic equation:   y² + 12y – 34 = 0. Step 5. Use the quadratic formula, y = [–b ± √(b² – 4ac)] / (2a), where a = 1, b = 12, and c = –34:   Discriminant: Δ = 12² – 4(1)(–34) = 144 + 136 = 280. Thus,   y = [–12 ± √280] / 2. Since √280 = 2√70, then   y = [–12 ± 2√70] / 2 = –6 ± √70. Step 6. Check the domain restrictions. For the logarithms we need:   a) y + 1 > 0 ⟹ y > –1,   b) 7 – 2y > 0 ⟹ y < 7/2 = 3.5. Evaluate both solutions:   • y = –6 – √70 is much less than –1 (and thus invalid).   • y = –6 + √70. Since √70 ≈ 8.37, we get y ≈ 2.37, which satisfies y > –1 and y < 3.5. Thus the unique solution is   y = –6 + √70. ───────────────────────────── Final Answers (i) x = –1/3 + (1/6)·log₂3   (with constants a = –1/3, b = 1/6) (ii) y = –6 + √70.