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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)) = ?
log2220+log220⋅log25−2log225log220+2log25=?
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
log2220+log220⋅log25−2log225log220+2log25=20+log220⋅log25−2(5)log220+2log25=10+log2(22.5)⋅log25log2(22.5)+2log25=10+(log222+log25)⋅log25log222+log25+2log25=10+(2+log25)⋅log252+3log25=10+2log25+log25⋅log252+3log25=10+2log25+(log25)22+3log25=10+2(2.3219)+(2.3219)22+3(2.3219)≈2.2346
Commented by mathdanisur last updated on 05/Aug/21
Thank You Ser
ThankYouSer

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