Question Number 4266 by Rasheed Soomro last updated on 06/Jan/16 $$\mathrm{Analyze}\:\mathrm{for}\:\:\mathrm{integer}\:\mathrm{solution}\:: \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{positive}\:\mathrm{integers}. \\ $$$$\mathrm{ax}^{\mathrm{a}} +\mathrm{by}^{\mathrm{b}} =\mathrm{cz}^{\mathrm{c}} \\ $$ Commented by Yozzii last updated on…
Question Number 4250 by Yozzii last updated on 06/Jan/16 $${Let}\:{u}=\frac{{ln}\left(\mathrm{3}−\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left(\mathrm{2}+\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${What}\:{is}\:{the}\:{value}\:{of}\:{u}?\: \\ $$$$ \\ $$$${Let}\:{k}=\frac{{ln}\left({x}−\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left({x}−\mathrm{1}+\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${For}\:{what}\:{values}\:{of}\:{x}\:{does} \\ $$$$\left({i}\right)\:{k}\:{converge}\:\left({ii}\right)\:{k}\:{diverge}? \\ $$ Commented by Yozzii…
Question Number 135305 by JulioCesar last updated on 12/Mar/21 Answered by MJS_new last updated on 12/Mar/21 $$\int\frac{{dx}}{{a}^{{x}} +{b}}= \\ $$$$\:\:\:\:\:\left[{t}={a}^{{x}} +{b}\:\rightarrow\:{dx}=\frac{{dt}}{{a}^{{x}} \mathrm{ln}\:{a}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\int\frac{{dt}}{{t}\left({t}−{b}\right)}=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\int\left(\frac{\mathrm{1}}{{t}−{b}}−\frac{\mathrm{1}}{{t}}\right){dt}= \\…
Question Number 4234 by Filup last updated on 04/Jan/16 $${n},\:{m}\in\mathbb{Z} \\ $$$${n}!={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)… \\ $$$$ \\ $$$$\mathrm{if}\:{n}!={m}\:\mathrm{where}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{value} \\ $$$${m}\:\mathrm{but}\:\mathrm{not}\:{n},\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{for}\:{n}? \\ $$ Commented by prakash jain last…
Question Number 135300 by bobhans last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)=? \\ $$ Answered by Olaf last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 4232 by Yozzii last updated on 03/Jan/16 Commented by Yozzii last updated on 04/Jan/16 $${I}'{m}\:{experiencing}\:{difficulty}\: \\ $$$${to}\:{solve}\:{the}\:\mathrm{2}{nd}\:{part}\:{without}\:{use}\:{of}\:{a}\:{theorem} \\ $$$${I}\:{posted}\:{sometime}\:{ago},\:{and}\: \\ $$$${generating}\:{functions}. \\ $$…
Question Number 135298 by Ahmed1hamouda last updated on 12/Mar/21 Answered by mr W last updated on 12/Mar/21 $$\mathrm{1} \\ $$$${z}={x}+{yi} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{x}+{yi}}=\frac{{x}−{yi}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\…
Question Number 135285 by mnjuly1970 last updated on 11/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{pre}−{calculus} \\ $$$$\:\:{if}\:\:\:\:\:\:\:{log}_{\mathrm{48}} ^{\mathrm{72}} +{log}_{\mathrm{54}} ^{\mathrm{12}} \:={k}\: \\ $$$$\:\:\:\:\:\:{then}\:\:{log}_{\:\mathrm{12}} ^{\:\mathrm{27}} \:\:=??? \\ $$$$\:\:\:\:\:\:\:\: \\ $$ Answered…
Question Number 69738 by Therealsincro last updated on 27/Sep/19 $${Sarah}\:{dances}\:{everyday}\:{of}\:{the}\:{week}, \\ $$$${including}\:{saturdays}\:{and}\:{sundays}. \\ $$$${In}\:{november}\:\mathrm{2018},\:{Sarah}\:{had}\:{to}\:{miss} \\ $$$${a}\:{few}\:{days}.\:{To}\:{control}\:{her}\:{absences} \\ $$$${she}\:{marks}\:{the}\:{day}\:{she}\:{missed}\:{class} \\ $$$${with}\:{a}\:\boldsymbol{{x}}\:{on}\:{the}\:{calendar}. \\ $$$${She}\:{marked}\:{the}\:\mathrm{5}{th},\:\mathrm{21}{st}\:{and}\:\mathrm{27}{th} \\ $$$${of}\:{november}. \\…
Question Number 69741 by ajfour last updated on 27/Sep/19 Commented by mr W last updated on 27/Sep/19 Commented by mr W last updated on 27/Sep/19…