Question Number 144387 by SOMEDAVONG last updated on 25/Jun/21
$$\mathrm{A}=\left(\left(\mathrm{log}_{\mathrm{2}} \mathrm{9}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} \mathrm{9}\right)}} ×\left(\sqrt{\mathrm{7}}\right)^{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{4}} \mathrm{7}}} =? \\ $$
Answered by liberty last updated on 25/Jun/21
$$\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{log}\:_{\mathrm{2}} \mathrm{9}\right)}=\mathrm{log}\:_{\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{9}\right)} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{A}=\left(\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{9}\right)^{\mathrm{log}\:_{\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{9}\right)} \left(\mathrm{2}\right)} \right)^{\mathrm{2}} ×\left(\mathrm{7}^{\mathrm{log}\:_{\mathrm{7}} \left(\mathrm{4}\right)} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{A}=\:\mathrm{4}×\sqrt{\mathrm{4}}\:=\:\mathrm{8}. \\ $$
Commented by SOMEDAVONG last updated on 25/Jun/21
$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$