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Given that log2093=c and log2097=f, express log(b extac3d) in terms of e and f.

Mathematics

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Answers

6 parts
a

3.6431\text{3.6431}

b

1.4057\text{1.4057}

c

2.9358\text{2.9358}

d

3\text{3}

e

1.9679\text{1.9679}

f

3\text{3}

Step-by-step solution

22 steps

Ili kutatua swali hili, tutatumia kanuni ya kubadilisha 'base' ya logariti, ambayo inasema log⁡ba=log⁡alog⁡b\log_b a = \frac{\log a}{\log b}, ambapo log⁡\log bila 'base' maalum inamaanisha 'base' 10. Pia tutatumia kanuni za logariti kama vile log⁡(xy)=log⁡x+log⁡y\log(xy) = \log x + \log y na log⁡(xn)=nlog⁡x\log(x^n) = n \log x.

Tunapewa:
log⁡2=0.36103\log 2 = 0.36103
log⁡3=0.47712\log 3 = 0.47712
log⁡5=0.69897\log 5 = 0.69897
log⁡7=0.8451\log 7 = 0.8451

Part a· 4 steps

log⁡2(18)\log_2 (18)

  1. Badilisha 'base' ya logariti.

    log⁡2(18)=log⁡18log⁡2\log_2 (18) = \frac{\log 18}{\log 2}

  2. Rahisisha log⁡18\log 18 kwa kutumia kanuni za logariti.

    log⁡18=log⁡(2×32)=log⁡2+log⁡(32)=log⁡2+2log⁡3\log 18 = \log (2 \times 3^2) = \log 2 + \log (3^2) = \log 2 + 2 \log 3

  3. Badilisha thamani ulizopewa na ukokotoe.

    log⁡18=0.36103+2(0.47712)=0.36103+0.95424=1.31527\log 18 = 0.36103 + 2(0.47712) = 0.36103 + 0.95424 = 1.31527 log⁡2(18)=1.315270.36103≈3.6431\log_2 (18) = \frac{1.31527}{0.36103} \approx 3.6431

  4. Jibu la mwisho.

    3.6431\boxed{\text{3.6431}}

Part b

1.4057\text{1.4057}

log⁡15(45)\log_{15} (45)

  1. Badilisha 'base' ya logariti.

    log⁡15(45)=log⁡45log⁡15\log_{15} (45) = \frac{\log 45}{\log 15}

  2. Rahisisha log⁡45\log 45 na log⁡15\log 15 kwa kutumia kanuni za logariti.

    log⁡45=log⁡(32×5)=2log⁡3+log⁡5\log 45 = \log (3^2 \times 5) = 2 \log 3 + \log 5 log⁡15=log⁡(3×5)=log⁡3+log⁡5\log 15 = \log (3 \times 5) = \log 3 + \log 5

  3. Badilisha thamani ulizopewa na ukokotoe.

    log⁡45=2(0.47712)+0.69897=0.95424+0.69897=1.65321\log 45 = 2(0.47712) + 0.69897 = 0.95424 + 0.69897 = 1.65321 log⁡15=0.47712+0.69897=1.17609\log 15 = 0.47712 + 0.69897 = 1.17609 log⁡15(45)=1.653211.17609≈1.4057\log_{15} (45) = \frac{1.65321}{1.17609} \approx 1.4057

  4. Jibu la mwisho.

    1.4057\boxed{\text{1.4057}}

Part c

2.9358\text{2.9358}

log⁡7(200)\log_7 (200)

  1. Badilisha 'base' ya logariti.

    log⁡7(200)=log⁡200log⁡7\log_7 (200) = \frac{\log 200}{\log 7}

  2. Rahisisha log⁡200\log 200 kwa kutumia kanuni za logariti.

    log⁡200=log⁡(23×52)=3log⁡2+2log⁡5\log 200 = \log (2^3 \times 5^2) = 3 \log 2 + 2 \log 5

  3. Badilisha thamani ulizopewa na ukokotoe.

    log⁡200=3(0.36103)+2(0.69897)=1.08309+1.39794=2.48103\log 200 = 3(0.36103) + 2(0.69897) = 1.08309 + 1.39794 = 2.48103 log⁡7(200)=2.481030.8451≈2.9358\log_7 (200) = \frac{2.48103}{0.8451} \approx 2.9358

  4. Jibu la mwisho.

    2.9358\boxed{\text{2.9358}}

Part d

3\text{3}

log⁡3(27)\log_3 (27)

  1. Andika 27 kama 'power' ya 3.

    27=3327 = 3^3

  2. Tumia kanuni ya logariti log⁡b(bn)=n\log_b (b^n) = n.

    log⁡3(27)=log⁡3(33)=3\log_3 (27) = \log_3 (3^3) = 3

  3. Jibu la mwisho.

    3\boxed{\text{3}}

Part e

1.9679\text{1.9679}

log⁡4(20)\log_4 (20)

  1. Badilisha 'base' ya logariti.

    log⁡4(20)=log⁡20log⁡4\log_4 (20) = \frac{\log 20}{\log 4}

  2. Rahisisha log⁡20\log 20 na log⁡4\log 4 kwa kutumia kanuni za logariti.

    log⁡20=log⁡(22×5)=2log⁡2+log⁡5\log 20 = \log (2^2 \times 5) = 2 \log 2 + \log 5 log⁡4=log⁡(22)=2log⁡2\log 4 = \log (2^2) = 2 \log 2

  3. Badilisha thamani ulizopewa na ukokotoe.

    log⁡20=2(0.36103)+0.69897=0.72206+0.69897=1.42103\log 20 = 2(0.36103) + 0.69897 = 0.72206 + 0.69897 = 1.42103 log⁡4=2(0.36103)=0.72206\log 4 = 2(0.36103) = 0.72206 log⁡4(20)=1.421030.72206≈1.9679\log_4 (20) = \frac{1.42103}{0.72206} \approx 1.9679

  4. Jibu la mwisho.

    1.9679\boxed{\text{1.9679}}

Part f

3\text{3}

log⁡4(64)\log_4 (64)

  1. Andika 64 kama 'power' ya 4.

    64=4364 = 4^3

  2. Tumia kanuni ya logariti log⁡b(bn)=n\log_b (b^n) = n.

    log⁡4(64)=log⁡4(43)=3\log_4 (64) = \log_4 (4^3) = 3

  3. Jibu la mwisho.

    3\boxed{\text{3}}
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