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25-1-log-2-a-1-5-1-log-4-a-3-a-0-5-Simplify-

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Question Number 5023 by love math last updated on 03/Apr/16
((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) Simplify
25−1+log2a−15−1−log4a−3a−0.5Simplify
Answered by Rasheed Soomro last updated on 05/Apr/16
((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) log_2 a^(−1) =((log_5 a^(−1) )/(log_5 2)) log_4 a=((log_5 a)/(log_5 4)) 25^(−1) =log_5 x⇒x=5^(25^(−1) ) 5^(−1) =log_5 x⇒x=5^5^(−1) ((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) )=((log_5 (5^(25^(−1) ) )+((log_5 a^(−1) )/(log_5 2)))/(log_5 (5^5^(−1) )−((log_5 a)/(log_5 4))))−(3/a^(−0.5) ) =((log_5 (5^(25^(−1) ) )log_5 2+log_5 a^(−1) )/(log_5 2))×((log_5 4)/(log_5 (5^5^(−1) )log_5 4−log_5 a))−(3/a^(−0.5) ) ((log_5 (5^(25^(−log_5 2) ) )+log_5 a^(−1) )/(log_5 (5^5^(−log_5 ^ 4) )−log_5 a))×((log_5 4)/(log_5 2))−(3/a^(−0.5) ) ((log_5 (5^(25^(−log_5 2) ) ×a^(−1) ))/(log_5 (5^5^(−log_5 ^ 4) ÷log_5 a)))×((log_5 4)/(log_5 2))−(3/a^(−0.5) ) Too complicated,I can′t continue.
25−1+log2a−15−1−log4a−3a−0.5log2a−1=log5a−1log52log4a=log5alog5425−1=log5x⇒x=525−15−1=log5x⇒x=55−125−1+log2a−15−1−log4a−3a−0.5=log5(525−1)+log5a−1log52log5(55−1)−log5alog54−3a−0.5=log5(525−1)log52+log5a−1log52×log54log5(55−1)log54−log5a−3a−0.5log5(525−log52)+log5a−1log5(55−log54)−log5a×log54log52−3a−0.5log5(525−log52×a−1)log5(55−log54\boldsymbol÷log5a)×log54log52−3a−0.5Toocomplicated,Ican′tcontinue.
Answered by Rasheed Soomro last updated on 05/Apr/16
((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) ((1−25log_2 a)/(25))×(5/(1−5log_4 a))−3(√a) ((1−25log_2 a)/(5−25log_4 a))−3(√a) ((1−25log_2 a)/(5−25(((log_2 a)/(log_2 4)))))−3(√a) ((1−25log_2 a)/(5−25(((log_2 a)/2))))−3(√a) ((1−25log_2 a)/(5−((25)/2)log_2 a))−3(√a)
25−1+log2a−15−1−log4a−3a−0.51−25log2a25×51−5log4a−3a1−25log2a5−25log4a−3a1−25log2a5−25(log2alog24)−3a1−25log2a5−25(log2a2)−3a1−25log2a5−252log2a−3a

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