Question Number 4582 by FilupSmith last updated on 08/Feb/16 $${S}=\mathrm{log}_{{a}} {i} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$${a}\in\mathbb{R} \\ $$$${Solve}\:{S} \\ $$$$ \\ $$$${extra}: \\ $$$$\mathrm{what}\:\mathrm{if}\:\:{a}\in\mathbb{C}? \\…
Question Number 70103 by Shamim last updated on 01/Oct/19 $$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$ Answered by mind is…
Question Number 4556 by FilupSmith last updated on 07/Feb/16 $$\mathrm{Prove}\:{i}^{{i}} ={e}^{−\pi/\mathrm{2}} ,\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$ Answered by Yozzii last updated on 07/Feb/16 $${i}=\mathrm{0}+{i}×\mathrm{1}={cos}\mathrm{0}.\mathrm{5}\pi+{isin}\mathrm{0}.\mathrm{5}={e}^{\mathrm{0}.\mathrm{5}\pi{i}} \\ $$$$\therefore{i}^{{i}}…
Question Number 4539 by love math last updated on 05/Feb/16 $${Two}\:{tractor},\:{working}\:{together},\: \\ $$$${have}\:{plowed}\:{the}\:{field}\:{for}\:\mathrm{48}\:{hours}. \\ $$$${If}\:{half}\:{of}\:{the}\:{field}\:{is}\:{plowed}\:{by}\: \\ $$$${one}\:{of}\:{them}\:{and}\:{then}\:{the}\:{remaining} \\ $$$${half}\:{to}\:{the}\:{other},\:{the}\:{work}\:{would} \\ $$$${be}\:{carried}\:{out}\:\mathrm{100}\:{hours}.\:{How}\:{many} \\ $$$${hours}\:{to}\:{plow}\:{the}\:{field}\:{if}\:{every}\:{tractor} \\ $$$${working}\:{separately}?…
Question Number 4524 by love math last updated on 05/Feb/16 $$\mathrm{3}^{\mathrm{3}{x}} +{x}^{\mathrm{2}} −\mathrm{9}>\mathrm{0} \\ $$ Commented by Yozzii last updated on 05/Feb/16 $$\mathrm{27}^{{x}} >\mathrm{9}−{x}^{\mathrm{2}} \:\:\:\left(\ast\right)…
Question Number 70040 by Shamim last updated on 30/Sep/19 $$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$ Commented by…
Question Number 4502 by #na last updated on 02/Feb/16 $$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}+\frac{\mathrm{6}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$ Answered by Rasheed Soomro last updated on 02/Feb/16 $$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}−\frac{\mathrm{6}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\…
Question Number 4492 by RasheedSindhi last updated on 01/Feb/16 $$ \\ $$$${By}\:{murging}\:{three}\:{sequences} \\ $$$$\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \:\&\:{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} ,…,{c}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\…
Question Number 4485 by Rasheed Soomro last updated on 31/Jan/16 $${By}\:\:{murging}\:\:{two}\:{sequences} \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…{a}_{{n}} \:\:\:{and}\:\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{b}_{\mathrm{2}}…
Question Number 135536 by islamo last updated on 13/Mar/21 $$\left(\boldsymbol{{u}}^{\boldsymbol{{v}}} \right)'=? \\ $$$${u}\:,{v}\:\:\:{tow}\:{fonctions} \\ $$ Answered by Olaf last updated on 13/Mar/21 $${u}^{{v}} \:=\:{e}^{\mathrm{ln}\left({u}^{{v}} \right)}…