Show that 2 log 5/8+ log 128/625 + log 5/2 = 0
Mathexpert:
It should be "2log 5/8 + log 128/625 + log 25/2"
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2 log 5/8 +log 128/625+log 5/2
2 {log5-log8} + log128-log625+log5-log2 use loga/b=loga-logb
2{log 5- log 2^3} + log 2^7-log5^4+log5-log2
2log5-6log2 +7 log2-4log5+log5-log2 use loga^b=bloga
-log5≠0
never equals to 0
2 {log5-log8} + log128-log625+log5-log2 use loga/b=loga-logb
2{log 5- log 2^3} + log 2^7-log5^4+log5-log2
2log5-6log2 +7 log2-4log5+log5-log2 use loga^b=bloga
-log5≠0
never equals to 0
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