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Solve for x in the equation; 1/2 log base 2 of 81 + log base 2 of (x2 - x/3) = 1

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4 parts
1

x=23 au x=−13x = \frac{2}{3} \text{ au } x = -\frac{1}{3}

2

H inapungua kwa 46.67%\text{H inapungua kwa } 46.67\%

Ksh. 62,389.52\text{Ksh. } 62,389.52

4
1.2166
5

−3−47-3 - 4\sqrt{7}

Step-by-step solution

17 steps

Part 1· 4 steps

Tatua xx katika mlinganyo: 12log⁡281+log⁡2(x2−x3)=1\frac{1}{2}\log_2 81 + \log_2 \left(x^2 - \frac{x}{3}\right) = 1

  1. Rahisisha neno la kwanza kwa kutumia sheria ya logariti (nlog⁡ba=log⁡ban)\left(n \log_b a = \log_b a^n\right).

    12log⁡281=log⁡28112=log⁡281=log⁡29\frac{1}{2}\log_2 81 = \log_2 81^{\frac{1}{2}} = \log_2 \sqrt{81} = \log_2 9

    Mlinganyo unakuwa:

    log⁡29+log⁡2(x2−x3)=1\log_2 9 + \log_2 \left(x^2 - \frac{x}{3}\right) = 1

  2. Unganisha maneno ya logariti kwa kutumia sheria ya kuzidisha (log⁡bM+log⁡bN=log⁡b(MN))\left(\log_b M + \log_b N = \log_b (MN)\right).

    log⁡2(9(x2−x3))=1\log_2 \left(9 \left(x^2 - \frac{x}{3}\right)\right) = 1 log⁡2(9x2−3x)=1\log_2 (9x^2 - 3x) = 1

  3. Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo (log⁡bA=C  ⟹  A=bC)\left(\log_b A = C \implies A = b^C\right).

    9x2−3x=219x^2 - 3x = 2^1 9x2−3x=29x^2 - 3x = 2

  4. Panga upya kuwa mlinganyo wa quadratic na utatue kwa xx.

    9x2−3x−2=09x^2 - 3x - 2 = 0

    Tumia fomula ya quadratic x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:
    Hapa a=9a=9, b=−3b=-3, c=−2c=-2.

    x=−(−3)±(−3)2−4(9)(−2)2(9)x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(9)(-2)}}{2(9)} x=3±9+7218x = \frac{3 \pm \sqrt{9 + 72}}{18} x=3±8118x = \frac{3 \pm \sqrt{81}}{18} x=3±918x = \frac{3 \pm 9}{18}

    Kuna suluhisho mbili:

    x1=3+918=1218=23x_1 = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3} x2=3−918=−618=−13x_2 = \frac{3 - 9}{18} = \frac{-6}{18} = -\frac{1}{3}

    Angalia kama suluhisho hizi ni halali kwa maneno ya logariti. Maneno ndani ya logariti lazima yawe chanya.
    Kwa x=23x = \frac{2}{3}: x2−x3=(23)2−2/33=49−29=29>0x^2 - \frac{x}{3} = \left(\frac{2}{3}\right)^2 - \frac{2/3}{3} = \frac{4}{9} - \frac{2}{9} = \frac{2}{9} > 0. Hii ni halali.
    Kwa x=−13x = -\frac{1}{3}: x2−x3=(−13)2−−1/33=19+19=29>0x^2 - \frac{x}{3} = \left(-\frac{1}{3}\right)^2 - \frac{-1/3}{3} = \frac{1}{9} + \frac{1}{9} = \frac{2}{9} > 0. Hii pia ni halali.

    x=23 au x=−13\boxed{x = \frac{2}{3} \text{ au } x = -\frac{1}{3}}

Part 2

H inapungua kwa 46.67%\text{H inapungua kwa } 46.67\%

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%.

  1. Andika mlinganyo wa tofauti.

    H=kVR2H = k \frac{V}{R^2}

    Ambapo kk ni kigezo cha uwiano.

  2. Eleza thamani mpya za V na R.

    V imeongezeka kwa 20%, kwa hivyo Vmpya=V+0.20V=1.2VV_{mpya} = V + 0.20V = 1.2V.
    R imeongezeka kwa 50%, kwa hivyo Rmpya=R+0.50R=1.5RR_{mpya} = R + 0.50R = 1.5R.

  3. Badilisha thamani mpya kwenye mlinganyo wa tofauti ili kupata HmpyaH_{mpya}.

    Hmpya=kVmpyaRmpya2=k1.2V(1.5R)2H_{mpya} = k \frac{V_{mpya}}{R_{mpya}^2} = k \frac{1.2V}{(1.5R)^2} Hmpya=k1.2V2.25R2H_{mpya} = k \frac{1.2V}{2.25R^2} Hmpya=1.22.25(kVR2)H_{mpya} = \frac{1.2}{2.25} \left(k \frac{V}{R^2}\right)

    Tunajua H=kVR2H = k \frac{V}{R^2}, kwa hivyo:

    Hmpya=1.22.25HH_{mpya} = \frac{1.2}{2.25} H Hmpya=0.5333...HH_{mpya} = 0.5333... H

  4. Kokotoa asilimia ya mabadiliko katika H.

    Asilimia ya mabadiliko =Hmpya−HH×100%= \frac{H_{mpya} - H}{H} \times 100\%.

    Asilimia ya mabadiliko=0.5333H−HH×100%\text{Asilimia ya mabadiliko} = \frac{0.5333H - H}{H} \times 100\% Asilimia ya mabadiliko=−0.4667HH×100%\text{Asilimia ya mabadiliko} = \frac{-0.4667H}{H} \times 100\% Asilimia ya mabadiliko=−0.4667×100%\text{Asilimia ya mabadiliko} = -0.4667 \times 100\% Asilimia ya mabadiliko=−46.67%\text{Asilimia ya mabadiliko} = -46.67\%

    H inapungua kwa 46.67%.

    H inapungua kwa 46.67%\boxed{\text{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?

  5. Kokotoa thamani baada ya mwaka wa kwanza.

    Thamani ya awali P=Ksh. 40,000P = \text{Ksh. } 40,000.
    Kiwango cha ongezeko mwaka wa 1 r1=12%=0.12r_1 = 12\% = 0.12.
    Thamani baada ya mwaka wa 1 A1=P(1+r1)=40,000(1+0.12)=40,000(1.12)=Ksh. 44,800A_1 = P(1 + r_1) = 40,000(1 + 0.12) = 40,000(1.12) = \text{Ksh. } 44,800.

  6. Kokotoa thamani baada ya mwaka wa pili.

    Kiwango cha ongezeko kwa miaka inayofuata r2=18%=0.18r_2 = 18\% = 0.18.
    Thamani baada ya mwaka wa 2 A2=A1(1+r2)=44,800(1+0.18)=44,800(1.18)=Ksh. 52,864A_2 = A_1(1 + r_2) = 44,800(1 + 0.18) = 44,800(1.18) = \text{Ksh. } 52,864.

  7. Kokotoa thamani baada ya mwaka wa tatu.

    Thamani baada ya mwaka wa 3 A3=A2(1+r2)=52,864(1.18)=Ksh. 62,389.52A_3 = A_2(1 + r_2) = 52,864(1.18) = \text{Ksh. } 62,389.52.

    Ksh. 62,389.52\boxed{\text{Ksh. } 62,389.52}

Part 41.2166

Panua (x2+1)5\left(\frac{x}{2} + 1\right)^5 hadi neno la nne, kisha tumia upanuzi huo kukadiria (1.04)5(1.04)^5 kwa usahihi wa sehemu 4 za desimali.

  1. Panua usemi wa binomial (x2+1)5\left(\frac{x}{2} + 1\right)^5 hadi neno la nne.

    Tunatumia fomula ya binomial (a+b)n=(n0)anb0+(n1)an−1b1+(n2)an−2b2+(n3)an−3b3+…(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + \dots
    Hapa a=1a = 1, b=x2b = \frac{x}{2}, na n=5n = 5. (Ni rahisi kuweka 1 kama 'a' ili nguvu za x ziwe zinaongezeka).

    (1+x2)5=(50)(1)5(x2)0+(51)(1)4(x2)1+(52)(1)3(x2)2+(53)(1)2(x2)3+…\left(1 + \frac{x}{2}\right)^5 = \binom{5}{0}(1)^5\left(\frac{x}{2}\right)^0 + \binom{5}{1}(1)^4\left(\frac{x}{2}\right)^1 + \binom{5}{2}(1)^3\left(\frac{x}{2}\right)^2 + \binom{5}{3}(1)^2\left(\frac{x}{2}\right)^3 + \dots =1⋅1⋅1+5⋅1⋅x2+10⋅1⋅x24+10⋅1⋅x38+…= 1 \cdot 1 \cdot 1 + 5 \cdot 1 \cdot \frac{x}{2} + 10 \cdot 1 \cdot \frac{x^2}{4} + 10 \cdot 1 \cdot \frac{x^3}{8} + \dots =1+5x2+10x24+10x38+…= 1 + \frac{5x}{2} + \frac{10x^2}{4} + \frac{10x^3}{8} + \dots =1+5x2+5x22+5x34+…= 1 + \frac{5x}{2} + \frac{5x^2}{2} + \frac{5x^3}{4} + \dots

  2. Linganisha upanuzi na (1.04)5(1.04)^5 ili kupata thamani ya xx.

    Tunataka kukadiria (1.04)5(1.04)^5.
    Tuna (1+x2)5\left(1 + \frac{x}{2}\right)^5.
    Kwa hivyo, 1+x2=1.041 + \frac{x}{2} = 1.04.

    x2=1.04−1\frac{x}{2} = 1.04 - 1 x2=0.04\frac{x}{2} = 0.04 x=0.04×2x = 0.04 \times 2 x=0.08x = 0.08

  3. Badilisha thamani ya xx kwenye maneno yaliyopanuliwa na ukokotoe makadirio.

    (1.04)5≈1+5(0.08)2+5(0.08)22+5(0.08)34(1.04)^5 \approx 1 + \frac{5(0.08)}{2} + \frac{5(0.08)^2}{2} + \frac{5(0.08)^3}{4} ≈1+0.402+5(0.0064)2+5(0.000512)4\approx 1 + \frac{0.40}{2} + \frac{5(0.0064)}{2} + \frac{5(0.000512)}{4} ≈1+0.20+0.0322+0.002564\approx 1 + 0.20 + \frac{0.032}{2} + \frac{0.00256}{4} ≈1+0.20+0.016+0.00064\approx 1 + 0.20 + 0.016 + 0.00064 ≈1.21664\approx 1.21664

    Kwa usahihi wa sehemu 4 za desimali:

    1.2166\boxed{1.2166}

Part 5

−3−47-3 - 4\sqrt{7}

Rahisisha na kurahisisha surd ifuatayo kabisa: 33+7−53−7\frac{3}{3+\sqrt{7}} - \frac{5}{3-\sqrt{7}}

  1. Rahisisha sehemu ya kwanza.

    Kuzidisha nambari na denomineta kwa kiambishi cha denomineta (3−7)(3-\sqrt{7}).

    33+7=3(3−7)(3+7)(3−7)\frac{3}{3+\sqrt{7}} = \frac{3(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}

    Tumia (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 kwa denomineta:

    =9−3732−(7)2=9−379−7=9−372= \frac{9 - 3\sqrt{7}}{3^2 - (\sqrt{7})^2} = \frac{9 - 3\sqrt{7}}{9 - 7} = \frac{9 - 3\sqrt{7}}{2}

  2. Rahisisha sehemu ya pili.

    Kuzidisha nambari na denomineta kwa kiambishi cha denomineta (3+7)(3+\sqrt{7}).

    53−7=5(3+7)(3−7)(3+7)\frac{5}{3-\sqrt{7}} = \frac{5(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})} =15+5732−(7)2=15+579−7=15+572= \frac{15 + 5\sqrt{7}}{3^2 - (\sqrt{7})^2} = \frac{15 + 5\sqrt{7}}{9 - 7} = \frac{15 + 5\sqrt{7}}{2}

  3. Toa sehemu mbili zilizorahisishwa.

    9−372−15+572\frac{9 - 3\sqrt{7}}{2} - \frac{15 + 5\sqrt{7}}{2}

    Kwa kuwa zina denomineta sawa, tunaweza kutoa nambari:

    =(9−37)−(15+57)2= \frac{(9 - 3\sqrt{7}) - (15 + 5\sqrt{7})}{2} =9−37−15−572= \frac{9 - 3\sqrt{7} - 15 - 5\sqrt{7}}{2}

    Unganisha maneno yanayofanana:

    =(9−15)+(−37−57)2= \frac{(9 - 15) + (-3\sqrt{7} - 5\sqrt{7})}{2} =−6−872= \frac{-6 - 8\sqrt{7}}{2}

    Gawanya kila neno kwa 2:

    =−62−872= \frac{-6}{2} - \frac{8\sqrt{7}}{2} =−3−47= -3 - 4\sqrt{7} −3−47\boxed{-3 - 4\sqrt{7}}
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