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Question-221760

Question Number 221760 by OmoloyeMichael last updated on 09/Jun/25 Answered by shunmisaki007 last updated on 09/Jun/25 $$\left(\mathrm{For}\:{x}>\mathrm{0}\:\mathrm{and}\:{x}\neq\mathrm{1}.\right) \\ $$$$\mathrm{log}_{{x}} \left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)^{\mathrm{log}_{\mathrm{3}} \left({x}\right)} =\mathrm{2} \\…

Is-i-an-imaginary-number-i-1-answer-with-logic-

Question Number 221697 by fantastic last updated on 09/Jun/25 $${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$ Answered by Ghisom last updated on 09/Jun/25 $$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\sqrt{\mathrm{i}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}}…

Question-221626

Question Number 221626 by fantastic last updated on 08/Jun/25 Answered by mr W last updated on 08/Jun/25 $${side}\:{length}\:{of}\:{square}\:=\mathrm{1} \\ $$$${shaded}\:{area}\:=\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}} \\…

Solve-for-x-7x-1-3-x-x-0-

Question Number 221620 by fantastic last updated on 08/Jun/25 $${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$ Answered by fantastic last updated on 08/Jun/25 $${or}\:\left(\mathrm{7}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}.{x}^{\frac{\mathrm{1}}{\mathrm{3}}}…