This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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I will solve questions 7, 8, and 9. For questions 7 and 8, where the base of the logarithm is not specified, I will assume a common logarithm (base 10), as is standard practice when "log" is written without a subscript. Question 7: Evaluate 2 5 - 3 2 + 32. Step 1: Apply the power rule of logarithms, a b = b^a. 2 5 = 5^2 = 25 3 2 = 2^3 = 8 The expression becomes: 25 - 8 + 32 Step 2: Apply the product and quotient rules of logarithms, a + b = (ab) and a - b = ((a)/(b)). Combine the terms: ((25)/(8)) + 32 ((25)/(8) × 32) Simplify the expression inside the logarithm: (25 × 4) 100 Step 3: Evaluate the logarithm. Since the base is 10, _10 100 asks "10 to what power equals 100?". 100 = 2 The value is 2. Question 8: Evaluate ( 32 + 128 - 729)/( 32 + 2 - 27). Step 1: Simplify the numerator using logarithm rules. Identify powers of common bases: 32 = 2^5, 128 = 2^7, 729 = 3^6. Numerator = 2^5 + 2^7 - 3^6 Apply the power rule a b = b^a: Numerator = 5 2 + 7 2 - 6 3 Combine like terms: Numerator = 12 2 - 6 3 Step 2: Simplify the denominator using logarithm rules. Identify powers of common bases: 32 = 2^5, 27 = 3^3. Denominator = 2^5 + 2 - 3^3 Apply the power rule: Denominator = 5 2 + 2 - 3 3 Combine like terms: Denominator = 6 2 - 3 3 Step 3: Simplify the entire fraction. Substitute the simplified numerator and denominator back into the fraction: (12 2 - 6 3)/(6 2 - 3 3) Factor out common terms from the numerator and denominator: (6(2 2 - 3))/(3(2 2 - 3)) Cancel out the common factor (2 2 - 3): (6)/(3) = 2 The value is 2. Question 9: Find the value of x that satisfies the equation (x+5) = 4 - (x+2). Step 1: Apply the quotient rule of logarithms to the right side of the equation. (x+5) = ((4)/(x+2)) Step 2: Equate the arguments of the logarithms. Since A = B implies A=B: x+5 = (4)/(x+2) Step 3: Solve the algebraic equation for x. Multiply both sides by (x+2): (x+5)(x+2) = 4 Expand the left side: x^2 + 2x + 5x + 10 = 4 x^2 + 7x + 10 = 4 Subtract 4 from both sides to set the quadratic equation to zero: x^2 + 7x + 6 = 0 Step 4: Factor the quadratic equation. Find two numbers that multiply to 6 and add to 7 (which are 1 and 6): (x+1)(x+6) = 0 This gives two potential solutions for x: x+1 = 0 x = -1 x+6 = 0 x = -6 Step 5: Check the validity of the solutions based on the domain of logarithms. The argument of a logarithm must be positive. For (x+5), we need x+5 > 0 x > -5. For (x+2), we need x+2 > 0 x > -2. Both conditions must be satisfied, so x must be greater than -2. Check x = -1: x+5 = -1+5 = 4 > 0 (Valid) x+2 = -1+2 = 1 > 0 (Valid) So, x = -1 is a valid solution. Check x = -6: x+5 = -6+5 = -1 (Not valid, as the argument must be positive) So, x = -6 is not a valid solution. The value of x that satisfies the equation is -1. Last free one today — make it count tomorrow, or type /upgrade for unlimited.