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Satu-satu dulu dijabarin karena keterbatasan media^ tanda pangkat
-) 1 / (^alog bc + 1)
= 1 / (^alog bc + ^alog a)
= 1 / (^alog (bc×a))
= 1 / ^alog abc
= ^(abc)log a
-) 1 / (^blog ac + 1)
= 1 / (^blog ac + ^blog b)
= 1 / ^blog abc
= ^(abc)log b
-) 1 / (^clog ab + 1)
= 1 / (^clog ab + ^clog c)
= 1 / (^clog abc)
= ^(abc)log c
maka jadinya
= ^(abc)log a + ^(abc)log b + ^(abc)log c
= ^(abc)log (abc)
= 1