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16-1-x-32-2x-1-128-2x-1-solve-

Question Number 163881 by abdurehime last updated on 11/Jan/22 $$\mathrm{16}^{\mathrm{1}−\mathrm{x}} \mathrm{32}^{\mathrm{2x}+\mathrm{1}} =\mathrm{128}^{\mathrm{2x}−\mathrm{1}} \:\:\:\:\mathrm{solve} \\ $$ Answered by Ar Brandon last updated on 11/Jan/22 $$\mathrm{2}^{\mathrm{4}\left(\mathrm{1}−{x}\right)+\mathrm{5}\left(\mathrm{2}{x}+\mathrm{1}\right)} =\mathrm{2}^{\mathrm{7}\left(\mathrm{2}{x}−\mathrm{1}\right)}…

Question-32810

Question Number 32810 by mondodotto@gmail.com last updated on 02/Apr/18 Commented by caravan msup abdo. last updated on 02/Apr/18 $${let}\:{use}\:{the}\:{ch}.\:\sqrt{\mathrm{3}}\:{x}\:=\mathrm{2}{sint} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{sint}\:\sqrt{\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} {t}}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{cost}\:{dt} \\…

f-f-n-2n-f-n-

Question Number 32743 by naka3546 last updated on 01/Apr/18 $${f}\:\left({f}\:\left({n}\right)\right)\:\:=\:\:\mathrm{2}{n} \\ $$$${f}\:\left({n}\right)\:\:=\:\:? \\ $$ Commented by abdo imad last updated on 03/Apr/18 $${if}\:{f}\:{is}\:{a}\:{polynomial}\:{fof}\left({x}\right)=\mathrm{2}{x}\:\Rightarrow{f}^{,} \left({f}\left({x}\right)\right){f}^{'} \left({x}\right)=\mathrm{2}\:\Rightarrow{y}^{,}…