Use logarithms correct to 4 decimal places to evaluate sqrt((0.3698 56)/(2.548)).

Here are the solutions to the problems. Problem 1: Use logarithms correct to 4 decimal places to evaluate sqrt((0.3698 56)/(2.548)). Step 1: Let N = sqrt((0.3698 56)/(2.548)). Taking the logarithm of both sides: N = ( (0.3698 56)/(2.548) )^1/2 N = (1)/(2) ( (0.3698) + ( 56^) - (2.548) ) Step 2: Find the logarithms of the individual terms. Using a calculator for 56^: 56^ ≈ 0.8290 (to 4 decimal places) Now, find the logarithms: (0.3698) = 1.5680 ( 56^) = (0.8290) = 1.9186 (2.548) = 0.4062 Step 3: Substitute the logarithm values into the equation for N. N = (1)/(2) ( 1.5680 + 1.9186 - 0.4062 ) First, add the logarithms in the numerator: 1.5680 + 1.9186 = (-1 + 0.5680) + (-1 + 0.9186) = -2 + 1.4866 = 1.4866 Now, subtract (2.548): 1.4866 - 0.4062 = (-1 + 0.4866) - 0.4062 = -1 + 0.0804 = 1.0804 So, N = (1)/(2) (1.0804) To divide the characteristic 1 by 2, we rewrite it as 2 + 1: N = (1)/(2) (-2 + 1.0804) N = -1 + 0.5402 = 1.5402 Step 4: Find the antilogarithm of 1.5402. Antilog(1.5402) = Antilog(-1 + 0.5402) Antilog(0.5402) ≈ 3.469 Therefore, N ≈ 0.3469. The value is: 0.3469 Problem 2: Solve the equation below: 7^2x - 8 × 7^x + 7 = 0. Step 1: Rewrite the equation using a substitution. Let y = 7^x. Then 7^2x = (7^x)^2 = y^2. The equation becomes a quadratic equation in terms of y: y^2 - 8y + 7 = 0 Step 2: Solve the quadratic equation for y. Factor the quadratic equation: (y-1)(y-7) = 0 This gives two possible values for y: y = 1 or y = 7 Step 3: Substitute back y = 7^x and solve for x. Case 1: y = 1 7^x = 1 Since 1 = 7^0, we have: 7^x = 7^0 x = 0 Case 2: y = 7 7^x = 7 Since 7 = 7^1, we have: 7^x = 7^1 x = 1 The solutions for x are: x=0, x=1 Problem 3: If $sqrt(14)sqrt(7)-sqrt(2) + sqrt(14)sqrt(7)+sqrt(2) = asqrt(7)+