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Category: Logarithms

log-10-x-log-10-x-8log-10-y-log-10-2-x-log-10-2-y-3-log-10-y-8log-10-x-log-10-y-log-10-2-x-log-10-2-y-0-find-x-amp-y-

Question Number 84459 by jagoll last updated on 13/Mar/20 $$\begin{cases}{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\frac{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{8log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{3}}\\{\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)+\frac{\mathrm{8log}_{\mathrm{10}} \left(\mathrm{x}\right)−\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{0}}\end{cases} \\…

log-x-1-6x-1-1-8-log-3-x-2-3-log-3-x-log-3-x-2-1-0-

Question Number 84051 by john santu last updated on 09/Mar/20 $$\frac{\mathrm{log}_{\left({x}−\mathrm{1}\right)} \:\left(\mathrm{6}{x}−\mathrm{1}\right)}{\left(\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)\right)^{\mathrm{3}} −\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\right)\left(\mathrm{log}_{\mathrm{3}} \:\left({x}−\mathrm{2}\right)−\mathrm{1}\right)}\:\geqslant\:\mathrm{0} \\ $$ Answered by john santu last updated…

x-2-log-5-x-x-5x-4-log-x-5-x-

Question Number 83786 by jagoll last updated on 06/Mar/20 $$\frac{{x}^{\mathrm{2}} }{\mathrm{log}_{\left(\mathrm{5}−{x}\right)} \:\left({x}\right)}\:\leqslant\:\left(\mathrm{5}{x}−\mathrm{4}\right)\:\mathrm{log}_{{x}} \:\left(\mathrm{5}−{x}\right)\: \\ $$ Answered by john santu last updated on 06/Mar/20 $${x}^{\mathrm{2}} \:\mathrm{log}_{{x}}…

Given-2-log-3-x-1-log-3-x-3-8-gt-0-have-the-solution-a-x-lt-b-what-is-b-

Question Number 83774 by jagoll last updated on 06/Mar/20 $$\mathrm{Given}\:\mathrm{2}\sqrt{\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}}\:−\:\mathrm{log}_{\mathrm{3}} \:{x}^{\mathrm{3}} \:+\mathrm{8}\:>\:\mathrm{0} \\ $$$${have}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{a}\:\leqslant\:{x}\:<\:{b}.\: \\ $$$${what}\:{is}\:{b}\:?\: \\ $$ Answered by john santu last updated…