Question Number 87637 by john santu last updated on 05/Apr/20 $$\sqrt[{\mathrm{4}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}+\mathrm{1}} }\:=\:\sqrt[{\mathrm{3}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}−\mathrm{2}} } \\ $$ Answered by TANMAY PANACEA. last updated on 05/Apr/20 $$\mid{x}−\mathrm{3}\mid^{\frac{{x}+\mathrm{1}}{\mathrm{4}}} =\mid{x}−\mathrm{3}\mid^{\frac{{x}−\mathrm{2}}{\mathrm{3}}}…
Question Number 153109 by peter frank last updated on 04/Sep/21 $$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{4}}} \\ $$ Commented by MJS_new last updated on 05/Sep/21 $$\mathrm{the}\:\mathrm{path}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{but}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{are}\:\mathrm{weird} \\ $$$${t}=\frac{{x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}}…
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Question Number 87194 by john santu last updated on 03/Apr/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\frac{\mid\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2}\mid}{\mathrm{x}−\mathrm{3}}\:<\:\mathrm{2}\: \\ $$ Commented by TANMAY PANACEA. last updated on 03/Apr/20 $${is}\:{it}\:\left(\mathrm{2}+{log}_{\mathrm{2}}…
Question Number 87125 by jagoll last updated on 03/Apr/20 Commented by john santu last updated on 03/Apr/20 $$\mathrm{cooll} \\ $$ Commented by john santu last…
Question Number 86515 by jagoll last updated on 29/Mar/20 $$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{x}\right)\:+\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{x}\right)\:=\:\mathrm{1}\: \\ $$$$\mathrm{x}\:=\: \\ $$ Commented by john santu last updated on 29/Mar/20 $$\Rightarrow\:\:^{\mathrm{2}}…
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Question Number 86339 by frempongfaustina24@gmail.com last updated on 28/Mar/20 $$\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} \\ $$$$ \\ $$$$ \\ $$ Answered by jagoll last updated on 28/Mar/20…
Question Number 86337 by frempongfaustina24@gmail.com last updated on 28/Mar/20 $$\left[\mathrm{4}\right. \\ $$ Commented by MJS last updated on 28/Mar/20 $$\mathrm{no}. \\ $$$$\mathrm{7} \\ $$ Terms…
Question Number 20626 by NECx last updated on 29/Aug/17 Answered by $@ty@m last updated on 30/Aug/17 $$\frac{\mathrm{2}}{{log}_{{a}} {x}}+\frac{\mathrm{1}}{{log}_{{a}} {ax}}+\frac{\mathrm{3}}{{log}_{{a}} {a}^{\mathrm{2}} {x}}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{log}_{{a}} {x}}+\frac{\mathrm{1}}{{log}_{{a}} {ax}}+\frac{\mathrm{1}}{{log}_{{a}}…