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Category: Algebra

solve-for-x-e-x-e-x-2-e-x-3-3-x-x-2-x-3-

Question Number 38587 by tawa tawa last updated on 27/Jun/18 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:+\:\mathrm{e}^{\mathrm{x}^{\mathrm{3}} } \:=\:\:\mathrm{3}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on…

Question-104115

Question Number 104115 by bobhans last updated on 19/Jul/20 Answered by nimnim last updated on 19/Jul/20 $$\frac{{xy}}{{x}+{y}}={a},\:\frac{{xz}}{{x}+{z}}={b},\:\frac{{yz}}{{y}+{z}}={c} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{a}},\:\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{b}},\:\:\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{c}} \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{bc}+{ac}+{ab}}{{abc}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{c}}=\frac{{bc}+{ac}+{ab}}{\mathrm{2}{abc}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{bc}+{ac}+{ab}}{\mathrm{2}{abc}}−\frac{\mathrm{1}}{{c}}…

Question-38570

Question Number 38570 by Sr@2004 last updated on 27/Jun/18 Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18 $${pls}\:{post}\:{question}\:{clearly}…{bhalo}\:{kore}\:{photo}\:{tolo} \\ $$$${tar}\:{por}\:{post}\:{karo}…{bangla}\:{language}\:{hole}\:{adubidha} \\ $$$${nai} \\ $$ Commented by…

Solve-log-r-8-log-3-p-5-i-r-p-11-ii-

Question Number 104086 by I want to learn more last updated on 19/Jul/20 $$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{log}_{\mathrm{r}} \mathrm{8}\:\:\:+\:\:\:\mathrm{log}_{\mathrm{3}} \mathrm{p}\:\:\:=\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:…..\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{r}\:\:\:+\:\:\mathrm{p}\:\:\:=\:\:\mathrm{11}\:\:\:\:\:\:\:\:\:…..\:\left(\mathrm{ii}\right) \\ $$ Commented by I want to…