2006-09-08 12:03:04 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You need the following 3 calculation rules for logarithms:
ln a + ln b = ln ab
ln a - ln b = ln (a/b)
a ln b = ln (b^a)
ln x + a ln y - b ln z =
ln x + ln (y^a) - ln (z^b) =
ln (x(y^a)/(z^b))
2006-09-08 12:14:15 · answer #1 · answered by mitch_online_nl 3 · 0⤊ 0⤋
the full element approximately logarithms is that we are able to maintain on with many regulations for them. it is the middle rule: ln a = b, e^b = a same element right here: ln (a million - 4y) = x e^x = a million - 4y e^x - a million = 4y y = (e^x - a million)/4
2016-12-12 05:02:10 · answer #2 · answered by karsten 4 · 0⤊ 0⤋
when you add two logs together you multiply the arguements and when you subtract you divide them. when you multiply a log by a number its the same as raising the arguement to that power. so...ln x +aln y - bln z = ln x + ln y^a - ln z^b = ln [ x * (y^a) / (z^b) ]
2006-09-08 12:08:47 · answer #3 · answered by Jake S 5 · 0⤊ 0⤋
ln xy to the power a divided by z to the power b
2006-09-08 12:51:37 · answer #4 · answered by mundi g 1 · 0⤊ 0⤋
ln x +a ln y - b ln z = ln x +ln (y^a) - ln (z^b) = ln (xy^a/z^b)
2006-09-08 12:07:16 · answer #5 · answered by Hex 2 · 1⤊ 2⤋
ln(x) + aln(y) - bln(z)
ln(x) + ln(y^a) - ln(z^b)
ln((x * y^a)/(z^b))
2006-09-08 12:31:47 · answer #6 · answered by Sherman81 6 · 0⤊ 0⤋
ln(x*y^a/z^b)
2006-09-08 12:10:37 · answer #7 · answered by bruinfan 7 · 0⤊ 0⤋