quotients, or powers?
2006-12-10 07:15:58 · 6 answers · asked by im_all_about_spades2003 1 in Science & Mathematics ➔ Mathematics
logy(xy)^7
= 7logxy
=7(logx + logy)
2006-12-10 07:17:16 · answer #1 · answered by alia_vahed 3 · 0⤊ 0⤋
log[288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e] = log 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e + log 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e - log 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e = 288fef27e03f3439e2bf4a983e0f9elog (x^6.y^15)/z^12 + 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9elog (x^6.y^15)/z^12 -288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e log (x^6.y^15)/z^12 so A= (x^6.y^15)/z^12 B= 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e C = -288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e 288fef27e03f3439e2bf4a983e0f9enegative 288fef27e03f3439e2bf4a983e0f9e288fef27e03f3439e2bf4a983e0f9e bc288fef27e03f3439e2bf4a983e0f9e the format given interior the question has +C (x^6.y^15)/z^12 wish it helped
2016-12-11 06:24:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋
OK,
log (xy)^7 = 7 log (xy) = 7 (log x + log y)
Products all gone. Done!
2006-12-10 07:16:45 · answer #3 · answered by Jerry P 6 · 0⤊ 0⤋
log (xy)^7 =
7*log (xy) = 7*log x + 7*log y
₢
2006-12-10 10:58:08 · answer #4 · answered by Luiz S 7 · 0⤊ 0⤋
log(xy)^7=7logxy=7(logx+logy)
2006-12-10 07:27:04 · answer #5 · answered by rollyben 1 · 0⤊ 0⤋
log(xy)^a = a [ log(x) + log(y)]
2006-12-10 07:17:52 · answer #6 · answered by Anonymous · 0⤊ 0⤋