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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You're on a roll — Step 1: Solve 11(i) for x and y. We are given two equations: 1) xy = 64 2) _x y + _y x = (5)/(2) For equation (2), use the change of base formula for logarithms, _y x = (1)/(_x y). Let k = _x y. Then the equation becomes: k + (1)/(k) = (5)/(2) Multiply by 2k to clear the denominators: 2k^2 + 2 = 5k Rearrange into a quadratic equation: 2k^2 - 5k + 2 = 0 Factor the quadratic equation: (2k-1)(k-2) = 0 This gives two possible values for k: 2k-1 = 0 k = (1)/(2) k-2 = 0 k = 2 Case 1: k = (1)/(2) Since k = _x y, we have _x y = (1)/(2). Convert to exponential form: y = x^1/2 = sqrt(x). Substitute this into the first equation xy = 64: x(sqrt(x)) = 64 x^1x^1/2 = 64 x^3/2 = 64 Raise both sides to the power of (2)/(3): x = 64^2/3 x = (4^3)^2/3 x = 4^2 x = 16 Now find y: y = sqrt(x) = sqrt(16) = 4 So, one solution is (x,y) = (16,4). Case 2: k = 2 Since k = _x y, we have _x y = 2. Convert to exponential form: y = x^2. Substitute this into the first equation xy = 64: x(x^2) = 64 x^3 = 64 Take the cube root of both sides: x = [3]64 x = 4 Now find y: y = x^2 = 4^2 = 16 So, another solution is (x,y) = (4,16). The values of x and y are x=16, y=4 or x=4, y=16. Step 2: Prove 11(ii) that if a^x = b^y = (ab)^xy, then x+y=1. Let a^x = b^y = (ab)^xy = K. From a^x = K, we have a = K^1/x. From b^y = K, we have b = K^1/y. Substitute these expressions for a and b into (ab)^xy = K: (K^1/x · K^1/y)^xy = K Using the exponent rule K^m · K^n = K^m+n: (K^(1)/(x) + (1)/(y))^xy = K Combine the exponents in the parenthesis: (K^(y+x)/(xy))^xy = K Using the exponent rule (K^m)^n = K^mn: K^(y+x)/(xy) · xy = K K^y+x = K^1 Assuming K ≠ 1 (which implies a ≠ 1 and b ≠ 1), we can equate the exponents: y+x = 1 Therefore, x+y=1 is proven. Step 3: Prove 11(iii) that AB = A + B and (A)/(B) = A - B. Proof for AB = A + B: Let _b A = x and _b B = y. By the definition of logarithms, this means A = b^x and B = b^y. Multiply A and B: AB = b^x · b^y Using the exponent rule b^m · b^n = b^m+n: AB = b^x+y Convert this exponential form back to logarithmic form: _b (AB) = x+y Substitute back the original definitions of x and y: _b (AB) = _b A + _b B When the base b is not specified, it is assumed to be a common base (e.g., 10 or e), so AB = A + B. Proof for (A)/(B) = A - B: Let _b A = x and _b B = y. By the definition of logarithms, this means A = b^x and B = b^y. Divide A by B: (A)/(B) = (b^x)/(b^y) Using the exponent rule (b^m)/(b^n) = b^m-n: (A)/(B) = b^x-y Convert this exponential form back to logarithmic form: _