2006-08-28 19:10:08 · 8 answers · asked by professional student 4 in Science & Mathematics ➔ Mathematics
why do you ask? guess what they say
2006-08-28 19:19:03 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First remember ln(a*b)= ln(a) + ln(b) so the first step is
ln(a) + ln(exp(-b*x)+c)
if exp(x) means e^x, then
ln(a) - b*x + c is the simplified expression.
If exp(x) means 10^x then ln(a) + ln(10)*[-b*x+c] is the result.
EDITED 8/28/06
2006-08-28 19:20:11 · answer #2 · answered by gp4rts 7 · 0⤊ 1⤋
remember ln(exp(1))=1
2006-08-28 19:13:59 · answer #3 · answered by female_lizzzzzard 3 · 0⤊ 0⤋
answer would be 0. via fact as multipy (x-a) (x-b) (x-c) ... (x-z) we can come for the time of (x-x) which would be equivalent to 0. And any huge type more suitable by making use of 0 will continuously provide 0 via fact the product.
2016-12-14 13:56:44 · answer #4 · answered by hust 4 · 0⤊ 0⤋
assuming * is multiplication,
ln [a.(e^-bx) + c] = ln [ a/e^bx + c] = ln [(a+c.e^bx)/e^bx)
= ln [a+ c.e^bx] - ln e^bx
= ln [a + c.e^bx] - bx
Don't reckon further simplification possible
2006-08-29 01:52:32 · answer #5 · answered by yasiru89 6 · 0⤊ 0⤋
ln aEXP-b^x + c
=ln a + ln e^(-b^x) +c
=ln a - b^x +c
2006-08-28 21:45:02 · answer #6 · answered by Nishu 1 · 0⤊ 0⤋
ln aEXP-b^x + c
=ln a + ln e^(-b^x) +c
=ln a - b^x +c
2006-08-28 19:17:35 · answer #7 · answered by M. Abuhelwa 5 · 0⤊ 1⤋
can not be simplified further
2006-08-28 19:19:17 · answer #8 · answered by oracle 5 · 0⤊ 0⤋