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E-log-a-c-b-log-ab-c-log-b-c-a-log-ab-c-




Question Number 139456 by SOMEDAVONG last updated on 27/Apr/21
E=log_a (√(c/b^(log_(ab) c) ))+log_b (√(c/a^(log_(ab) c) ))
E=logacblogabc+logbcalogabc
Answered by bemath last updated on 27/Apr/21
E = log _a (√(c/c^(log _(ab) b) )) + log _b (√(c/c^(log _(ab) a) ))  E=(1/2)[ log _a (c^(1−log _(ab) b) )+log _b (c^(1−log _(ab) a) )]  E=(1/2) [ log _a (c^(log _(ab) ((1/a))) )+ log _b (c^(log _(ab) ((1/b))) )]
E=logacclogabb+logbcclogabaE=12[loga(c1logabb)+logb(c1logaba)]E=12[loga(clogab(1a))+logb(clogab(1b))]

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