Solve the value of x. i) log₂(x-1) + log₂x = 6; ii) log₃(x+1) + log₃(x+2) = 3.

Hapa kuna hatua za kutatua milinganyo ya logariti: i) _2 (x-1) + _2 x = 6 Step 1: Tumia sheria ya logariti _b M + _b N = _b (M · N) kuunganisha maneno. _2 ((x-1)x) = 6 _2 (x^2 - x) = 6 Step 2: Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia _b A = C b^C = A . 2^6 = x^2 - x 64 = x^2 - x Step 3: Panga upya mlinganyo kuwa fomu ya quadratic ax^2 + bx + c = 0. x^2 - x - 64 = 0 Step 4: Tumia fomula ya quadratic x = -b ± sqrt(b^2 - 4ac)2a kutatua kwa x. Hapa a=1, b=-1, c=-64. x = -(-1) ± sqrt((-1)^2 - 4(1)(-64))2(1) x = 1 ± sqrt(1 + 256)2 x = 1 ± sqrt(257)2 Step 5: Angalia masharti ya logariti. Kwa _2 (x-1) na _2 x kufafanuliwa, tunahitaji x-1 > 0 x > 1 na x > 0. Hivyo, x lazima iwe kubwa kuliko 1. x_1 = 1 + sqrt(257)2 ≈ (1 + 16.03)/(2) ≈ 8.515 x_2 = 1 - sqrt(257)2 ≈ (1 - 16.03)/(2) ≈ -7.515 Thamani ya x_2 si kubwa kuliko 1, kwa hivyo si suluhisho halali. Suluhisho la kwanza ni: x = 1 + sqrt(257)2 ii) _3 (x+1) + _3 (x+2) = 3 Step 1: Tumia sheria ya logariti _b M + _b N = _b (M · N) kuunganisha maneno. _3 ((x+1)(x+2)) = 3 _3 (x^2 + 2x + x + 2) = 3 _3 (x^2 + 3x + 2) = 3 Step 2: Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia _b A = C b^C = A . 3^3 = x^2 + 3x + 2 27 = x^2 + 3x + 2 Step 3: Panga upya mlinganyo kuwa fomu ya quadratic ax^2 + bx + c = 0. x^2 + 3x + 2 - 27 = 0 x^2 + 3x - 25 = 0 Step 4: Tumia fomula ya quadratic x = -b ± sqrt(b^2 - 4ac)2a kutatua kwa x. Hapa a=1, b=3, c=-25. x = -3 ± sqrt(3^2 - 4(1)(-25))2(1) x = -3 ± sqrt(9 + 100)2 x = -3 ± sqrt(109)2 Step 5: Angalia masharti ya logariti. Kwa _3 (x+1) na _3 (x+2) kufafanuliwa, tunahitaji x+1 > 0 x > -1 na x+2 > 0 x > -2. Hivyo, x lazima iwe kubwa kuliko -1. x_1 = -3 + sqrt(109)2 ≈ (-3 + 10.44)/(2) ≈ 3.72 x_2 = -3 - sqrt(109)2 ≈ (-3 - 10.44)/(2) ≈ -6.72 Thamani ya x_2 si kubwa kuliko -1, kwa hivyo si suluhisho halali. Suluhisho la pili ni: x = -3 + sqrt(109)2 Tuma swali linalofuata 📸