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Swali la 1: Tatua x katika mlinganyo: (1)/(2)_2 81 + _2 (x^2 - (x)/(3)) = 1 Hatua ya 1: Rahisisha neno la kwanza kwa kutumia sheria ya logariti (n _b a = _b a^n). (1)/(2)_2 81 = _2 81^(1)/(2) = _2 sqrt(81) = _2 9 Mlinganyo unakuwa: _2 9 + _2 (x^2 - (x)/(3)) = 1 Hatua ya 2: Unganisha maneno ya logariti kwa kutumia sheria ya kuzidisha (_b M + _b N = _b (MN)). _2 (9 (x^2 - (x)/(3))) = 1 _2 (9x^2 - 3x) = 1 Hatua ya 3: Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo (_b A = C A = b^C). 9x^2 - 3x = 2^1 9x^2 - 3x = 2 Hatua ya 4: Panga upya kuwa mlinganyo wa quadratic na utatue kwa x. 9x^2 - 3x - 2 = 0 Tumia fomula ya quadratic x = -b ± sqrt(b^2 - 4ac)2a: Hapa a=9, b=-3, c=-2. x = -(-3) ± sqrt((-3)^2 - 4(9)(-2))2(9) x = 3 ± sqrt(9 + 72)18 x = 3 ± sqrt(81)18 x = (3 ± 9)/(18) Kuna suluhisho mbili: x_1 = (3 + 9)/(18) = (12)/(18) = (2)/(3) x_2 = (3 - 9)/(18) = (-6)/(18) = -(1)/(3) Angalia kama suluhisho hizi ni halali kwa maneno ya logariti. Maneno ndani ya logariti lazima yawe chanya. Kwa x = (2)/(3): x^2 - (x)/(3) = ((2)/(3))^2 - (2/3)/(3) = (4)/(9) - (2)/(9) = (2)/(9) > 0. Hii ni halali. Kwa x = -(1)/(3): x^2 - (x)/(3) = (-(1)/(3))^2 - (-1/3)/(3) = (1)/(9) + (1)/(9) = (2)/(9) > 0. Hii pia ni halali. x = (2)/(3) au x = -(1)/(3) Swali la 2: H inatofautiana moja kwa moja na V na kinyume na mraba wa R. Tafuta asilimia ya mabadiliko katika H ikiwa V imeongezeka kwa 20% na R kwa 50%. Hatua ya 1: Andika mlinganyo wa tofauti. H = k (V)/(R^2) Ambapo k ni kigezo cha uwiano. Hatua ya 2: Eleza thamani mpya za V na R. V imeongezeka kwa 20%, kwa hivyo V_mpya = V + 0.20V = 1.2V. R imeongezeka kwa 50%, kwa hivyo R_mpya = R + 0.50R = 1.5R. Hatua ya 3: Badilisha thamani mpya kwenye mlinganyo wa tofauti ili kupata H_mpya. H_mpya = k V_mpyaR_mpya^2 = k (1.2V)/((1.5R)^2) H_mpya = k (1.2V)/(2.25R^2) H_mpya = (1.2)/(2.25) (k (V)/(R^2)) Tunajua H = k (V)/(R^2), kwa hivyo: H_mpya = (1.2)/(2.25) H H_mpya = 0.5333... H Hatua ya 4: Kokotoa asilimia ya mabadiliko katika H. Asilimia ya mabadiliko = H_mpya - HH × 100\%. Asilimia ya mabadiliko = (0.5333H - H)/(H) × 100\% Asilimia ya mabadiliko = (-0.4667H)/(H) × 100\% Asilimia ya mabadiliko = -0.4667 × 100\% Asilimia ya mabadiliko = -46.67\% H inapungua kwa 46.67%. H inapungua kwa 46.67\% Swali la 3: David alinunua kipande cha ardhi kwa Ksh.40,000 ambacho kiliongezeka thamani kwa 12% p.a katika mwaka wa kwanza na 18% p.a kwa miaka iliyofuata. Ikiwa aliuza ardhi baada ya miaka mitatu, alipata kiasi gani? Hatua ya 1: Kokotoa thamani baada ya mwaka wa kwanza. Thamani ya awali P = Ksh. 40,000. Kiwango cha ongezeko mwaka wa 1 r_1 = 12\% = 0.12. Thamani baada ya mwaka wa 1 A_1 = P(1 + r_1) = 40,000(1 + 0.12) = 40,000(1.12) = Ksh. 44,800. Hatua ya 2: Kokotoa thamani baada ya mwaka wa pili. Kiwango cha ongezeko kwa miaka inayofuata r_2 = 18\% = 0.18. Thamani baada ya mwaka wa 2 A_2 = A_1(1 + r_2) = 44,800(1 + 0.18) = 44,800(1.18) = Ksh. 52,864. Hatua ya 3: Kokotoa thamani baada ya mwaka wa tatu. Thamani baada ya mwaka wa 3 A_3 = A_2(1 + r_2) = 52,864(1.18) = Ksh. 62,389.52. Ksh. 62,389.52 Swali la 4: Panua ((x)/(2) + 1)^5 hadi neno la nne, kisha tumia upanuzi huo kukadiria (1.04)^5 kwa usahihi wa sehemu 4 za desimali. Hatua ya 1: Panua usemi wa binomial ((x)/(2) + 1)^5 hadi neno la nne. Tunatumia fomula ya binomial (a+b)^n = n0a^n b^0 + n1a^n-1 b^1 + n2a^n-2 b^2 + n3a^n-3 b^3 + Hapa a = 1, b = (x)/(2), na n = 5. (Ni rahisi kuweka 1 kama 'a' ili nguvu za x ziwe zinaongezeka). (1 + (x)/(2))^5 = 50(1)^5((x)/(2))^0 + 51(1)^4((x)/(2))^1 + 52(1)^3((x)/(2))^2 + 53(1)^2((x)/(2))^3 + = 1 · 1 · 1 + 5 · 1 · (x)/(2) + 10 · 1 · (x^2)/(4) + 10 · 1 · (x^3)/(8) + = 1 + (5x)/(2) + (10x^2)/(4) + (10x^3)/(8) + = 1 + (5x)/(2) + (5x^2)/(2) + (5x^3)/(4) + Hatua ya 2: Linganisha upanuzi na (1.04)^5 ili kupata thamani ya x. Tunataka kukadiria (1.04)^5. Tuna (1 + (x)/(2))^5. Kwa hivyo, 1 + (x)/(2) = 1.04. (x)/(2) = 1.04 - 1 (x)/(2) = 0.04 x = 0.04 × 2 x = 0.08 Hatua ya 3: Badilisha thamani ya x kwenye maneno yaliyopanuliwa na ukokotoe makadirio. (1.04)^5 ≈ 1 + (5(0.08))/(2) + (5(0.08)^2)/(2) + (5(0.08)^3)/(4) ≈ 1 + (0.40)/(2) + (5(0.0064))/(2) + (5(0.000512))/(4) ≈ 1 + 0.20 + (0.032)/(2) + (0.00256)/(4) ≈ 1 + 0.20 + 0.016 + 0.00064 ≈ 1.21664 Kwa usahihi wa sehemu 4 za desimali: 1.2166 Swali la 5: Rahisisha na kurahisisha surd ifuatayo kabisa: (3)/(3+sqrt(7)) - (5)/(3-sqrt(7)) Hatua ya 1: Rahisisha sehemu ya kwanza. Kuzidisha nambari na denomineta kwa kiambishi cha denomineta (3-sqrt(7)). (3)/(3+sqrt(7)) = 3(3-sqrt(7))(3+sqrt(7))(3-sqrt(7)) Tumia (a+b)(a-b) = a^2 - b^2 kwa denomineta: = 9 - 3sqrt(7)3^2 - (sqrt(7))^2 = 9 - 3sqrt(7)9 - 7 = 9 - 3sqrt(7)2 Hatua ya 2: Rahisisha sehemu ya pili. Kuzidisha nambari na denomineta kwa kiambishi cha denomineta (3+sqrt(7)). (5)/(3-sqrt(7)) = 5(3+sqrt(7))(3-sqrt(7))(3+sqrt(7)) = 15 + 5sqrt(7)3^2 - (sqrt(7))^2 = 15 + 5sqrt(7)9 - 7 = 15 + 5sqrt(7)2 Hatua ya 3: Toa sehemu mbili zilizorahisishwa. 9 - 3sqrt(7)2 - 15 + 5sqrt(7)2 Kwa kuwa zina denomineta sawa, tunaweza kutoa nambari: = (9 - 3sqrt(7)) - (15 + 5sqrt(7))2 = 9 - 3sqrt(7) - 15 - 5sqrt(7)2 Unganisha maneno yanayofanana: = (9 - 15) + (-3sqrt(7) - 5sqrt(7))2 = -6 - 8sqrt(7)2 Gawanya kila neno kwa 2: = (-6)/(2) - 8sqrt(7)2 = -3 - 4sqrt(7) -3 - 4sqrt(7) Tuma swali linalofuata 📸