Question Number 172601 by Mikenice last updated on 29/Jun/22
Answered by Rasheed.Sindhi last updated on 30/Jun/22
![•log_2 (((75)/(16)))−log_2 [(((((25)/(81)))^(3/4) (((25)/(81)))^(1/4) )/((((125)/(405)))^(1/2) ))]^2 +(1/3)(log_2 2^(15) +log_2 3^(−15) ) [(((((25)/(81)))^(3/4) (((25)/(81)))^(1/4) )/((((125)/(405)))^(1/2) ))]^2 =[(((((25)/(81)))^((3/4)+(1/4)) )/((((125)/(405)))^(1/2) ))]^2 =(((((25)/(81)))^2 )/(((((125)/(405)))^(1/2) )^2 ))=(5^4 /3^8 )×((3^4 ×5)/5^3 )=(5^2 /3^4 ) •log_2 (((75)/(16)))−log_2 ((5^2 /3^4 ))+(1/3)(log_2 2^(15) +log_2 3^(−15) ) •log_2 (((75)/(16))÷(5^2 /3^4 ))+(1/3)(log_2 2^(15) ×3^(−15) ) •log_2 (((75)/(16))×(3^4 /5^2 ))+(1/3)(log_2 (2^(15) ×3^(−15) ) •log_2 ((3^5 /2^4 ))+log_2 ((2^5 /3^5 )) •log_2 ((3^5 /2^4 )×(2^5 /3^5 )) •log_2 (2)=1](Q172705.png)
$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{75}}{\mathrm{16}}\right)−\mathrm{log}_{\mathrm{2}} \left[\frac{\left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\mathrm{3}/\mathrm{4}} \left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\mathrm{1}/\mathrm{4}} }{\left(\frac{\mathrm{125}}{\mathrm{405}}\right)^{\mathrm{1}/\mathrm{2}} }\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{15}} +\mathrm{log}_{\mathrm{2}} \mathrm{3}^{−\mathrm{15}} \right) \\ $$$$\left[\frac{\left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\mathrm{3}/\mathrm{4}} \left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\mathrm{1}/\mathrm{4}} }{\left(\frac{\mathrm{125}}{\mathrm{405}}\right)^{\mathrm{1}/\mathrm{2}} }\right]^{\mathrm{2}} =\left[\frac{\left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\frac{\mathrm{125}}{\mathrm{405}}\right)^{\mathrm{1}/\mathrm{2}} }\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\left(\frac{\mathrm{25}}{\mathrm{81}}\right)^{\mathrm{2}} }{\left(\left(\frac{\mathrm{125}}{\mathrm{405}}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{5}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{8}} }×\frac{\mathrm{3}^{\mathrm{4}} ×\mathrm{5}}{\mathrm{5}^{\mathrm{3}} }=\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{4}} } \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{75}}{\mathrm{16}}\right)−\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{4}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{15}} +\mathrm{log}_{\mathrm{2}} \mathrm{3}^{−\mathrm{15}} \right) \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{75}}{\mathrm{16}}\boldsymbol{\div}\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{4}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{15}} ×\mathrm{3}^{−\mathrm{15}} \right) \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{75}}{\mathrm{16}}×\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{5}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{15}} ×\mathrm{3}^{−\mathrm{15}} \right)\right. \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{3}^{\mathrm{5}} }{\mathrm{2}^{\mathrm{4}} }\right)+\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{2}^{\mathrm{5}} }{\mathrm{3}^{\mathrm{5}} }\right) \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{3}^{\mathrm{5}} }{\mathrm{2}^{\mathrm{4}} }×\frac{\mathrm{2}^{\mathrm{5}} }{\mathrm{3}^{\mathrm{5}} }\right) \\ $$$$\bullet\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}\right)=\mathrm{1} \\ $$