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(²log1/63+²log7)³log8/5log16√2×⁴log1/625
Jawabannya adalah 2/3
Konsep
^a log b + ^a log c = ^a log bc
^a log b + ^b log c = ^a log c
^a log a = 1
^a log a^m = m
^aⁿ log a^m = m/n
Asumsikan soal hitunglah[ (²log1/63+²log7)³log8] / [^5log16√2×⁴log1/625]
[ (²log1/63+²log7)³log8] / [^5log16√2×⁴log1/625]
=[ (²log1/63(7))³log2³] / [^5log2⁴(2)^(½)×⁴log1/5⁴]
=[ (²log1/9)³log2³] / [^5log2^(4½)×⁴log5^(-4)]
=[ (²log1/3²)³log2³] / [(-4)x(4½)⁴log5 x^5log2]
=[ (²log3^(-2)) x³log2³] / [(-4)x(4½)x⁴log5 x^5log2]
= [-2x 3 x ²log3 x ³log2] / [-18x ⁴log5 x^5log2]
=[-6 x ²log2 ] / [-18x ⁴log2]
= [-6 x ²log2 ] / [-18x ^2²log2]
= [-6 x ²log2 ] / [-18x (½)²log2]
= – 6 / (-18x½)
= – 6 / – 9
= 2/3
Jadi jawabnya adalah 2/3