??
thnx for ur help =]
2006-08-26 11:21:56 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
e^(1+ ln x) = e^1 * e^(ln x) = e * x
That is as simple as it can get.
2006-08-26 11:26:37 · answer #1 · answered by Vincent G 7 · 3⤊ 0⤋
Its the same as (e^1)(e^lnx). Since e^lnx=x, your problem simplifies to xe.
2006-08-26 11:26:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋
e^(1 + ln(x)) = e^1 * e^(lnx) = e^1 * x = e * x
2006-08-26 12:06:39 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
e^(1+lnx) = e^1 * e^(lnx) = e * x
thats because e^(ln(anything))=anything
2006-08-26 11:29:32 · answer #4 · answered by QA Guy 3 · 1⤊ 0⤋
e^(1+lnX) = e^1 * e^lnX = e * X = eX
2006-08-26 11:27:19 · answer #5 · answered by lippy19850528 3 · 1⤊ 1⤋
e^ vital -lnx might desire to be written as e^(vital -ln(x) dx) =e^(-vital ln(x) dx) Now, by using factors, vital (lnx)dx = xln(x)-x+C_1, the place C_1 is the consistent of integration. So e^(-vital ln(x) dx) = e^(-xln(x)-x+C_1)=e^(-xln(x))*e^(-x)*e^C... =C_2x^(-x)*e^(-x), the place C_2=e^C_1 is a non-0 consistent, and e^(-xln(x))=x^(-x). ideal style is C_2(xe)^(-x).
2016-11-05 21:24:00 · answer #6 · answered by ? 4 · 0⤊ 0⤋
well by exponent rules,
e^ (1+ ln x) = (e^1).(e^ln x)
= e . x
(e is the coefficient and x is the variable)
2006-08-27 02:59:16 · answer #7 · answered by yasiru89 6 · 0⤊ 0⤋
sure! e^(1+lnx)= e^(1+lnx)
that was easy!
2006-08-26 11:25:34 · answer #8 · answered by abcdefg_female 2 · 1⤊ 2⤋
e*x is probably it
2006-08-26 11:28:35 · answer #9 · answered by Anonymous · 1⤊ 0⤋
no idea sorry
2006-08-26 11:24:00 · answer #10 · answered by Anonymous · 0⤊ 2⤋