nilai dari ^7log 16 . ^8log 49 - ^5log 125 per ^6log 4 + ^6log 9

Hai Auffa, terima kasih sudah bertanya di Roboguru. Kakak bantu jawab ya :) Jawaban : (7/6) - (3/2)(^2log(3)) + (2)(^6log(3)) Ingat sifat bentuk logaritma berikut ini : ^(a^m)log(b^n) = (n/m) ^alog(b) ^alog(a) = 1 ^alog(b) = 1/(^blog(a)) ^alog(b) . ^blog(c) = ^alog(c) ^alog(b) + ^alog(c) = ^alog(b.c) ^7log(16) . ^8log(49) - (^5log(125) / ^6log(4)) + ^6log(9) = ^7log(2^4) . ^(2^3)log(7^2) - (^5log(5^3) / ^6log(2^2)) + ^6log(3^2) = (4)(^7log(2)) . (2/3)(^2log(7)) - ((3)(^5log(5)) / (2)(^6log(2))) + (2)(^6log(3)) = (4)(2/3)(^7log(2) . ^2(log(7))) - (3/2)(^2log(6)) + (2)(^6log(3)) = (8/3)(1) - (3/2)(^2log(2.3)) + (2)(^6log(3)) = (8/3) - (3/2)(^2log(2) + ^2log(3)) + (2)(^6log(3)) = (8/3) - (3/2)(1 + ^2log(3)) + (2)(^6log(3)) = (8/3) - (3/2) - (3/2)(^2log(3)) + (2)(^6log(3)) = (16/6) - (9/6) - (3/2)(^2log(3)) + (2)(^6log(3)) = (7/6) - (3/2)(^2log(3)) + (2)(^6log(3)) Jadi, nilai dari logaritma di atas adalah (7/6) - (3/2)(^2log(3)) + (2)(^6log(3)). Semoga membantu ya. Semangat Belajar! :)