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

$$7\mathrm{C}={\log }_2\frac{1}{6}+{\log }_{3}27$$$$$$$$\mathrm{Solve}$$

Answered by BaliramKumar last updated on 27/Sep/22

$$7C=-{log}_2\left(6\right)+3{log}_3\left(3\right)$$$$7C=-{log}_2\left(2\right)-{log}_2\left(3\right)+3$$$$7C=-1-{log}_2\left(3\right)+3$$$$7C=2-{log}_2\left(3\right)$$$$C=\frac{2}{7}-\frac{{log}_2\left(3\right)}{7}$$$$C=\frac{{log}_2\left(4\right)-{log}_2\left(3\right)}{7}$$$$C=\frac{{\log }_2\left(\frac{4}{3}\right)}{7}$$$$$$

Answered by Frix last updated on 28/Sep/22

$${\log }_2\frac{1}{6}=-{\log }_{2}6=-\frac{\ln 6}{\ln 2}=-\frac{\ln 3+\ln 2}{\ln 2}=-1-\frac{\ln 3}{\ln 2}$$$${\log }_{3}27={\log }_3{3}^3=3$$$$7C=2-\frac{\ln 3}{\ln 2}$$$$C=\frac{2}{7}-\frac{\ln 3}{7\ln 2}(=\frac{2}{7}-\frac{1}{7}{\log }_{2}3)$$

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