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Question Number 149410 by mathdanisur last updated on 05/Aug/21

((log_2 2^(20) + log_2 20 ∙ log_2 5 - 2 log_2 2^5 )/(log_2 20 + 2 log_2 5)) = ?

$$\frac{{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{20}} \:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} }{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 05/Aug/21

((log_2 2^(20) + log_2 20 ∙ log_2 5 - 2 log_2 2^5 )/(log_2 20 + 2 log_2 5)) =((20 + log_2 20 ∙ log_2 5 - 2(5))/(log_2 20 + 2 log_2 5)) =((10 + log_2 (2^2 .5)∙ log_2 5 )/(log_2 (2^2 .5) + 2 log_2 5)) =((10 +(log_2 2^2 +log_2 5)∙ log_2 5 )/(log_2 2^2 +log_2 5 + 2 log_2 5)) =((10 +(2+log_2 5)∙ log_2 5 )/(2+ 3 log_2 5)) =((10 +2log_2 5+log_2 5∙ log_2 5 )/(2+ 3 log_2 5)) =((10 +2log_2 5+(log_2 5)^2 )/(2+ 3 log_2 5)) =((10 +2(2.3219)+(2.3219)^2 )/(2+ 3(2.3219)))≈2.2346

$$\frac{{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{20}} \:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} }{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\: \\ $$$$=\frac{\mathrm{20}\:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\left(\mathrm{5}\right)}{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\:{log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}} .\mathrm{5}\right)\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:}{{log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}} .\mathrm{5}\right)\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\left({log}_{\mathrm{2}} \mathrm{2}^{\mathrm{2}} +{log}_{\mathrm{2}} \mathrm{5}\right)\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:}{{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{2}} +{log}_{\mathrm{2}} \mathrm{5}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\left(\mathrm{2}+{log}_{\mathrm{2}} \mathrm{5}\right)\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:}{\mathrm{2}+\:\mathrm{3}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\mathrm{2}{log}_{\mathrm{2}} \mathrm{5}+{log}_{\mathrm{2}} \mathrm{5}\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:}{\mathrm{2}+\:\mathrm{3}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\mathrm{2}{log}_{\mathrm{2}} \mathrm{5}+\left({log}_{\mathrm{2}} \mathrm{5}\right)^{\mathrm{2}} \:}{\mathrm{2}+\:\mathrm{3}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:\: \\ $$$$=\frac{\mathrm{10}\:+\mathrm{2}\left(\mathrm{2}.\mathrm{3219}\right)+\left(\mathrm{2}.\mathrm{3219}\right)^{\mathrm{2}} \:}{\mathrm{2}+\:\mathrm{3}\left(\mathrm{2}.\mathrm{3219}\right)}\approx\mathrm{2}.\mathrm{2346} \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thank You Ser

$${Thank}\:{You}\:\mathrm{Ser} \\ $$

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