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log-3-x-1-2-x-

Question Number 189174 by sciencestudentW last updated on 12/Mar/23 $${log}_{\mathrm{3}} \left({x}+\mathrm{1}\right)=\mathrm{2}\:\:\:\:\:\:\:\:;\:\:\:{x}=? \\ $$ Answered by HeferH last updated on 12/Mar/23 $$\mathrm{log}\:_{{b}} {a}\:=\:{x} \\ $$$${b}^{{x}} \:=\:{a}…

Question-189140

Question Number 189140 by 073 last updated on 12/Mar/23 Answered by Ar Brandon last updated on 12/Mar/23 $${x}>\mathrm{0}\:\wedge\:{x}\geqslant\mathrm{1}\:\Rightarrow{x}\geqslant\mathrm{1} \\ $$$$\mathrm{4}−\mathrm{ln}{x}\geqslant\mathrm{3}\sqrt{\mathrm{ln}{x}}\:,\:{t}^{\mathrm{2}} =\mathrm{ln}{x} \\ $$$$\mathrm{4}−{t}^{\mathrm{2}} \geqslant\mathrm{3}{t}\:\Rightarrow\:{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{4}\:\leqslant\:\mathrm{0}…

sir-TINKUTARA-what-is-the-problem-with-brackets-in-red-color-1-1-2-5x-1-2-5x-

Question Number 123608 by malwan last updated on 26/Nov/20 $${sir}\:{TINKUTARA} \\ $$$${what}\:{is}\:{the}\:{problem}\:{with}\:{brackets} \\ $$$${in}\:{red}\:{color}\:? \\ $$$$\left.\mathrm{1}\right)\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \:\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \\ $$ Commented by JDamian last…

if-a-1-i-2-find-a-1943-

Question Number 123607 by mohammad17 last updated on 26/Nov/20 $${if}\:{a}=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:\:{find}\:{a}^{\mathrm{1943}} \:\:\:\:? \\ $$ Answered by Dwaipayan Shikari last updated on 26/Nov/20 $${a}\:=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{{i}} \\ $$$${a}^{\mathrm{1943}} =\left({i}\right)^{\frac{\mathrm{1943}}{\mathrm{2}}}…

0-2-x-4-x-dx-

Question Number 123600 by aurpeyz last updated on 26/Nov/20 $$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\sqrt{\frac{{x}}{\mathrm{4}−{x}}}{dx} \\ $$ Answered by Dwaipayan Shikari last updated on 26/Nov/20 $$\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\frac{{x}}{\mathrm{4}−{x}}}\:{dx}…

Question-123601

Question Number 123601 by aurpeyz last updated on 26/Nov/20 Commented by aurpeyz last updated on 26/Nov/20 $${pls}\:{explain}\:{to}\:{me}\:{if}\:{i}\:{still}\:{need}\:{to} \\ $$$${find}\:{the}\:{class}\:{boundary}\:{since}\:{the}\: \\ $$$${class}\:{interval}\:{is}\:{already}\:{continous}. \\ $$$${thank}\:{you} \\ $$…