Question Number 84826 by jagoll last updated on 16/Mar/20 $$\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{2}}\:+\:\mathrm{log}_{\mathrm{3}} \:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{10}}\:=\:\mathrm{2} \\ $$ Commented by john santu last updated on 16/Mar/20 $$\mathrm{x}\:=\:\mathrm{1} \\…
Question Number 84674 by jagoll last updated on 15/Mar/20 $$\frac{\mathrm{6}−\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{3}+\mathrm{2log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right)}\:<\:\mathrm{2} \\ $$ Commented by jagoll last updated on 15/Mar/20 $$\left(\mathrm{i}\right)\:\mathrm{x}\:\neq\:\mathrm{0}\: \\…
Question Number 84459 by jagoll last updated on 13/Mar/20 $$\begin{cases}{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\frac{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{8log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{3}}\\{\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)+\frac{\mathrm{8log}_{\mathrm{10}} \left(\mathrm{x}\right)−\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{0}}\end{cases} \\…
Question Number 84051 by john santu last updated on 09/Mar/20 $$\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} \\ $$ Answered by john santu last updated…
Question Number 84000 by byaw last updated on 08/Mar/20 Commented by Kunal12588 last updated on 08/Mar/20 $${log}\:\sqrt[{\mathrm{3}}]{\mathrm{0}.\mathrm{0005957}}\:{should}\:{be}\:−{ve}\:? \\ $$ Answered by MJS last updated on…
Question Number 18365 by aplus last updated on 19/Jul/17 Answered by mrW1 last updated on 19/Jul/17 $$\mathrm{x}=\mathrm{6} \\ $$$$\mathrm{y}=\mathrm{2} \\ $$ Commented by mrW1 last…
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 $${x}^{\mathrm{2}} \:\mathrm{log}_{{x}}…
Question Number 83774 by jagoll last updated on 06/Mar/20 $$\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…
Question Number 83706 by jagoll last updated on 05/Mar/20 $$\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{10}^{\mathrm{x}} \:=\:\mathrm{25}^{\mathrm{x}} \\ $$$$\mathrm{x}\:=\:? \\ $$ Answered by john santu last updated on 05/Mar/20 Terms…
Question Number 18153 by aplus last updated on 16/Jul/17 Commented by prakash jain last updated on 16/Jul/17 $$\mathrm{4}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{500}^{\mathrm{500}} \\ $$$$\mathrm{2}{x}\mathrm{log}\:\mathrm{4}=\mathrm{log}\:\mathrm{2}+\mathrm{500log}\:\mathrm{500} \\ $$$$\mathrm{3}=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{500log}\:\mathrm{500}}{\mathrm{2log}\:\mathrm{4}}\approx\mathrm{1120}.\mathrm{973} \\ $$…