Question Number 121876 by oustmuchiya@gmail.com last updated on 12/Nov/20 Commented by liberty last updated on 12/Nov/20 $$\Rightarrow\:\sqrt{\mathrm{x}−\mathrm{6}}\:=\:\mathrm{7}−\sqrt{\mathrm{x}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{6}\:=\:\mathrm{49}−\mathrm{14}\sqrt{\mathrm{x}−\mathrm{1}}\:+\:\mathrm{x}−\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{14}\sqrt{\mathrm{x}−\mathrm{1}}\:=\:\mathrm{54}\:\Rightarrow\sqrt{\mathrm{x}−\mathrm{1}}\:=\:\frac{\mathrm{27}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{1}\:=\:\frac{\mathrm{27}^{\mathrm{2}} }{\mathrm{49}}\:;\:\mathrm{x}\:=\:\frac{\mathrm{27}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}}…
Question Number 187406 by Fridunatjan08 last updated on 16/Feb/23 $$\mathrm{3}^{{x}} .\mathrm{8}^{\frac{{x}}{{x}+\mathrm{2}}} =\mathrm{6}{ln}\mathrm{0}−{log}\mathrm{10} \\ $$ Commented by Fridunatjan08 last updated on 16/Feb/23 $${can}\:{someone}\:{please}\:{help}\:{me}? \\ $$ Commented…
Question Number 121783 by Ar Brandon last updated on 11/Nov/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{satisfying}\: \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{2x}+\mathrm{1}<\mathrm{2log}_{\mathrm{2}} \left(\mathrm{x}+\mathrm{3}\right)\:\mathrm{is}\:\_\_\_. \\ $$ Answered by TANMAY PANACEA last updated on 11/Nov/20 $$…
Question Number 121778 by Ar Brandon last updated on 11/Nov/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{inequality} \\ $$$$\mathrm{3}^{\left(\mathrm{5}/\mathrm{2}\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{12}−\mathrm{3x}\right)} −\mathrm{3}^{\mathrm{log}_{\mathrm{2}} \mathrm{x}} >\mathrm{83}\:\mathrm{is}\:\_\_\_\_\: \\ $$ Commented by 676597498 last updated on…
Question Number 121387 by Ar Brandon last updated on 07/Nov/20 $$\frac{\left(\mathrm{81}\right)^{\mathrm{1}/\mathrm{log}_{\mathrm{5}} \mathrm{9}} +\mathrm{3}^{\mathrm{3}/\mathrm{log}_{\sqrt{\mathrm{6}}} \mathrm{3}} }{\mathrm{409}}\left[\left(\sqrt{\mathrm{7}}\right)^{\mathrm{2}/\mathrm{log}_{\mathrm{25}} \mathrm{7}} −\left(\mathrm{125}\right)^{\mathrm{log}_{\mathrm{25}} \mathrm{6}} \right] \\ $$ Answered by 675480065 last…
Question Number 55806 by Tawa1 last updated on 04/Mar/19 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\mathrm{x}^{\mathrm{log}_{\mathrm{3}} \mathrm{2}} \:\:=\:\:\sqrt{\mathrm{x}}\:\:+\:\:\mathrm{1} \\ $$ Answered by MJS last updated on 04/Mar/19 $$\mathrm{ln}\:{x}^{\mathrm{log}_{\mathrm{3}} \:\mathrm{2}} =\mathrm{ln}\:{x}^{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}} =\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\mathrm{ln}\:{x}\:=…
Question Number 121278 by Ar Brandon last updated on 06/Nov/20 $$\mathrm{The}\:\mathrm{domaine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\sqrt{\mathrm{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{13}\right)}\right)^{−\mathrm{1}} \:\mathrm{is}; \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 121201 by Ar Brandon last updated on 05/Nov/20 Commented by Ar Brandon last updated on 06/Nov/20 Commented by MJS_new last updated on 06/Nov/20…
Question Number 121175 by Ar Brandon last updated on 05/Nov/20 Answered by MJS_new last updated on 05/Nov/20 $${x}<\mathrm{1} \\ $$$${g}\left({x}\right)<\mathrm{1} \\ $$ Commented by Ar…
Question Number 121138 by Ar Brandon last updated on 05/Nov/20 Answered by TANMAY PANACEA last updated on 05/Nov/20 $${x}^{{x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} } =\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{{x}+{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} } \\ $$$${x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}}…