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 $${x}^{{x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} } =\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{{x}+{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} } \\ $$$${x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 121100 by Ar Brandon last updated on 05/Nov/20 Answered by TANMAY PANACEA last updated on 05/Nov/20 $$\frac{{logx}}{{x}\left({y}+{z}−{x}\right)}=\frac{{logy}}{{y}\left({z}+{x}−{y}\right)}=\frac{{logz}}{{z}\left({x}+{y}−{z}\right)}={k} \\ $$$${logx}={kx}\left({y}+{z}−{x}\right) \\ $$$${logy}={ky}\left({z}+{x}−{y}\right) \\ $$$${logz}={kz}\left({x}+{y}−{z}\right)…
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Question Number 121002 by Ar Brandon last updated on 04/Nov/20 Answered by MJS_new last updated on 04/Nov/20 $$ \\ $$ Commented by Ar Brandon last…
Question Number 120927 by Ar Brandon last updated on 04/Nov/20 Answered by 675480065 last updated on 04/Nov/20 $$\mathrm{domain}:\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{x}>\mathrm{3}/\mathrm{8}\:\cap\:\mathrm{x}>\mathrm{0} \\ $$$$\mathrm{hence}\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{3}/\mathrm{4}\:<\mathrm{0}…