e to the power of (-x) = 0.471
Find x.
I remember it has something to do with In, but cannot remember how to move down the negative sign?
2006-10-20 22:51:19 · 14 answers · asked by a V a 4 in Science & Mathematics ➔ Mathematics
e^(-x) = 0.471
Taking ln on both sides,
-x ln e = ln 0.471
-x = ln 0.471 (because ln e = 1)
x = -(ln 0.471)
x = 0.753
2006-10-20 22:59:38 · answer #1 · answered by Anonymous · 1⤊ 0⤋
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2016-05-22 07:17:05 · answer #2 · answered by Anonymous · 0⤊ 0⤋
e to the power of (-x)=1/e to the power of (x)
1/e to the power of (x)=0.471
therefore e to the power of (x)= 1/0.471 which is equal to 2.1231.
i could solve only this much sorry bcos im not a math student. Im a microbiologist.
2006-10-20 23:03:11 · answer #3 · answered by green 2 · 0⤊ 0⤋
Whoa that looks pretty difficult. Sorry I can't help. Besides, I'm not one for doing math at 5:00 in the morning.
2006-10-20 22:53:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Divide both sides by 1 to the power of (-x).I think.
2006-10-20 22:56:33 · answer #5 · answered by jad D 1 · 0⤊ 0⤋
ans: x = - (ln 0.471)
since
-x = ln 0.471
so bring the negative sign over.
2006-10-20 23:00:13 · answer #6 · answered by Anonymous · 0⤊ 0⤋
first take ln of e i.e .-xln(e) = ln (0.471)
it will become -xlne=(0.471)
as a law lne = 1
so -x= ln(0.471)
use calculator to calc ln(0.0471)
now give me my pionts
2006-10-20 23:01:25 · answer #7 · answered by sunny k 2 · 0⤊ 0⤋
e raised to pwer(-x)= 0.471
e^-x = 0.471 (^ -> raised to power)
we can write it as:
(log(0.417)/log e) = -x
(log e=2.718)
solving the above equation u'll get x(hope you know how to use log)
tell me is this help ful for u?
2006-10-20 23:12:05 · answer #8 · answered by indian 2 · 0⤊ 0⤋
e^-x=0.471 take log base e of both sides
-xln(e)=ln 0,471, ln(e) =1 because it is base e
x=-ln0.471 = -(-0.753)=0.753
2006-10-21 00:02:35 · answer #9 · answered by sydney m 2 · 0⤊ 0⤋
Take ln of both sides
-x=ln0.47
so x= -ln0.47
Don't have my calculator on me, sure you do.
2006-10-20 22:57:35 · answer #10 · answered by palmerdrury 1 · 0⤊ 0⤋