7C-log-2-1-6-log-3-27-Solve-

Question Number 176895 by Mastermind last updated on 27/Sep/22

7 C = \log_{2} \frac{1}{6} + \log_{3} 27

Solve

Answered by BaliramKumar last updated on 27/Sep/22

7 C = - {l o g}_{2} (6) + 3 {l o g}_{3} (3)

7 C = - {l o g}_{2} (2) - {l o g}_{2} (3) + 3

7 C = - 1 - {l o g}_{2} (3) + 3

7 C = 2 - {l o g}_{2} (3)

C = \frac{2}{7} - \frac{{l o g}_{2} (3)}{7}

C = \frac{{l o g}_{2} (4) - {l o g}_{2} (3)}{7}

C = \frac{\log_{2} (\frac{4}{3})}{7}

Answered by Frix last updated on 28/Sep/22

\log_{2} \frac{1}{6} = - \log_{2} 6 = - \frac{\ln 6}{\ln 2} = - \frac{\ln 3 + \ln 2}{\ln 2} = - 1 - \frac{\ln 3}{\ln 2}

\log_{3} 27 = \log_{3} 3^{3} = 3

7 C = 2 - \frac{\ln 3}{\ln 2}

C = \frac{2}{7} - \frac{\ln 3}{7 \ln 2} (= \frac{2}{7} - \frac{1}{7} \log_{2} 3)

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