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Question-171667

Question Number 171667 by Mastermind last updated on 19/Jun/22 Answered by mindispower last updated on 19/Jun/22 $${ln}\left({y}\right)={ln}\left({x}\right){ln}\left({cos}\left({x}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}{log}_{\mathrm{4}} \left({x}\right)\right) \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{1}}{{x}}{ln}\left({cos}\left({x}\right)\right)−{tg}\left({x}\right){ln}\left({x}\right)+\frac{\mathrm{2}{x}−\mathrm{5}+\frac{\mathrm{3}}{{xln}\left(\mathrm{4}\right)}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}{log}_{\mathrm{4}} \left({x}\right)\right)} \\ $$$${y}'=\left({cos}\left({x}\right)\right)^{{ln}\left({x}\right)}…

Question-171666

Question Number 171666 by Mastermind last updated on 19/Jun/22 Commented by kaivan.ahmadi last updated on 19/Jun/22 $$\mathrm{3}^{{x}} +{log}_{\mathrm{2}} {x}=\mathrm{10}\Rightarrow{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{3}^{{x}} } +{log}_{\mathrm{2}} {x}=\mathrm{10}\Rightarrow \\ $$$${x}×\mathrm{2}^{\mathrm{3}^{{x}}…