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\frac { 1 } { 1 + \operatorname { log } _ { b } a + \operatorname { log } _ { b } c } + \frac { 1 } { 1 + \operatorname { log } _ { c } a + \operatorname { log } _ { c } b } + \frac { 1 } { 1 + \operatorname { log } _ { a } b + \operatorname { log } _ { a } c } |
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Answer» using the identity loga to the base b = loga/logb we get = 1/(1+loga/logb +logc/logb) + 1/(1+loga/logc+logb/logc) +1/(1+logb/loga +logc/loga) = logb/(loga+logb+logc) + logc/(loga+logb+logc) +loga/(loga+logb+logc) = (loga+logb+logc)/(loga+logb+logc) = 1. |
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