condense the expression to the logarithm of a single quantity
lnx-4[ln(x+2)+ln(x-2)]
2006-12-14 07:26:37 · 5 answers · asked by dude4587 2 in Science & Mathematics ➔ Mathematics
lnx-4[ln(x+2)+ln(x-2)]
= lnx - 4ln[(x + 2)(x - 2)] {As logM + log N = logMN}
= lnx - 4ln[(x² - 4)]
= lnx - ln[(x² - 4)^4] {As nlogM = log[M^n]}
= ln[x/(x² - 4)^4] {As logM - logN = log[M/N]}
2006-12-14 07:33:00 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
lnx -4 ln (x+2) *(x-2) since ln a +ln b = ln ab
lnx -4 ln (x^2-4)
and since a * ln x = lnx ^ a
lnx - ln (x^2-4)^4 since ln a -ln b = ln a/b
finally ln x (x^2-4)^-4 = ln (x/((x^2-4)^4))
2006-12-14 15:34:03 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
lnx-4[ln(x+2)+ln(x-2)]
=lnx- ln(x^2-4)^4
ln (x/(x^2-4)^4)
hint:
ln (a * b)=ln a+ln b
ln ( a / b )= ln a- ln b
2006-12-14 15:32:47 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
lnx-4[ln(x+2)+ln(x-2)]
ln x-ln (x+2)^4-ln (x-2)^4
ln(x/(x^2-4)^4)
2006-12-14 15:37:35 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
lnx-4(ln(x+2)+ln(x-2))=
lnx-4(ln(x^2-4))=
ln(x/(x^2-4)^4)
2006-12-14 15:35:16 · answer #5 · answered by Anonymous · 0⤊ 0⤋