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Category: Algebra

Question-102002

Question Number 102002 by Jamshidbek2311 last updated on 06/Jul/20 Commented by mr W last updated on 06/Jul/20 $${please}\:{don}'{t}\:{post}\:{the}\:{same}\:{question} \\ $$$${again}\:{and}\:{again}.\:{if}\:{you}\:{have}\:{doubt}, \\ $$$${give}\:{comment}\:{in}\:{the}\:{old}\:{thread},\:{don}'{t} \\ $$$${open}\:{a}\:{new}\:{thread}. \\…

Suppose-x-and-y-are-real-such-that-log-4-x-log-6-y-log-9-x-y-find-y-x-

Question Number 167539 by Tawa11 last updated on 19/Mar/22 $$\mathrm{Suppose}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that},\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{x}\:\:\:=\:\:\:\mathrm{log}_{\mathrm{6}} \mathrm{y}\:\:\:=\:\:\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{x}\:+\:\mathrm{y}\right), \\ $$$$\mathrm{find}\:\:\:\frac{\mathrm{y}}{\mathrm{x}} \\ $$ Answered by Rasheed.Sindhi last updated on 19/Mar/22 $$\:\mathrm{log}_{\mathrm{4}}…

a-x-2-b-x-1-4-9-4a-b-2-

Question Number 167489 by mathlove last updated on 18/Mar/22 $${a}+\sqrt{{x}}=\mathrm{2} \\ $$$${b}+\sqrt[{\mathrm{4}}]{{x}}=\mathrm{9}\:\:\:\:\:\:\:\:\:\overset{\:\:\:\:\:\:\left(\mathrm{4}{a}−{b}^{\mathrm{2}} \right)=?} {\:} \\ $$ Answered by Rasheed.Sindhi last updated on 18/Mar/22 $${a}+\sqrt{{x}}=\mathrm{2}…\left({i}\right) \\…

a-2-4a-2-0-a-2-a-4-4-12a-2-

Question Number 167488 by mathlove last updated on 18/Mar/22 $${a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{2}=\mathrm{0}\:\:\:\:\:\:\:\:\:{a}\succ\mathrm{2} \\ $$$$\sqrt{\frac{{a}^{\mathrm{4}} +\mathrm{4}}{\mathrm{12}{a}^{\mathrm{2}} }}=? \\ $$ Commented by cortano1 last updated on 18/Mar/22 $$\:\:\:\left(\mathrm{a}−\mathrm{2}\right)^{\mathrm{2}}…

Please-help-i-cannot-solve-problems-such-as-x-4-1-x-5-1-or-x-n-1-n-N-

Question Number 101921 by Farruxjano last updated on 05/Jul/20 $$\boldsymbol{{Please}},\:\boldsymbol{{help}}\:\:\boldsymbol{{i}}\:\:\boldsymbol{{cannot}}\:\boldsymbol{{solve}}\:\boldsymbol{{problems}} \\ $$$$\boldsymbol{{such}}\:\boldsymbol{{as}}:\:\boldsymbol{{x}}^{\mathrm{4}} =\mathrm{1},\:\boldsymbol{{x}}^{\mathrm{5}} =\mathrm{1},\:\boldsymbol{{or}}\:\boldsymbol{{x}}^{\boldsymbol{{n}}} =\mathrm{1}\:\boldsymbol{{n}}\in\mathbb{N}… \\ $$ Commented by prakash jain last updated on 05/Jul/20…