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 updated on 07/Nov/20
$$\left(\frac{\mathrm{9}^{\mathrm{log}_{\mathrm{9}} \mathrm{25}} +\mathrm{3}^{\mathrm{log}_{\mathrm{3}} \mathrm{6}\sqrt{\mathrm{6}}} }{\mathrm{409}}\right)\left[\mathrm{7}^{\mathrm{log}_{\mathrm{7}} \mathrm{25}} −\mathrm{5}^{\mathrm{log}_{\mathrm{5}} \mathrm{6}\sqrt{\mathrm{6}}} \right] \\ $$$$=\left(\frac{\mathrm{25}+\mathrm{6}\sqrt{\mathrm{6}}}{\mathrm{409}}\right)\left(\mathrm{25}−\mathrm{6}\sqrt{\mathrm{6}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{409}}\right)\left(\mathrm{25}^{\mathrm{2}} −\mathrm{36}\left(\mathrm{6}\right)\right)=\frac{\mathrm{409}}{\mathrm{409}}=\mathrm{1} \\ $$
Answered by liberty last updated on 07/Nov/20
$$\left[\left(\mathrm{9}\right)^{\mathrm{log}\:_{\mathrm{9}} \left(\mathrm{5}\right)^{\mathrm{2}} } +\mathrm{3}^{\mathrm{log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{6}}\right)^{\mathrm{3}} } \right]\:\left[\:\left(\mathrm{7}\right)^{\mathrm{log}\:_{\mathrm{7}} \left(\mathrm{25}\right)} −\left(\mathrm{6}\right)^{\mathrm{log}\:_{\mathrm{25}} \left(\mathrm{125}\right)} \:\right]×\frac{\mathrm{1}}{\mathrm{409}} \\ $$$$=\:\left[\:\mathrm{25}+\mathrm{6}\sqrt{\mathrm{6}}\:\right]\left[\mathrm{25}−\mathrm{6}\sqrt{\mathrm{6}}\:\right]×\frac{\mathrm{1}}{\mathrm{409}} \\ $$$$=\:\left(\:\mathrm{625}−\mathrm{216}\right)×\frac{\mathrm{1}}{\mathrm{409}}\:=\:\mathrm{1} \\ $$$$\mathrm{625}−\mathrm{216} \\ $$$$\mathrm{409}.\mathrm{0} \\ $$
Commented by Ar Brandon last updated on 07/Nov/20
Cool. Thanks Sir