i'm drawing a blank on my precalc knowledge but im doing a physics problem and have
e^(-.3093t)= 0.03
i know i gotta use natural logs but im not totally sure how.
i need to solve for t
2007-04-10 13:08:30 · 5 answers · asked by sabresfan58 1 in Science & Mathematics ➔ Mathematics
ln(e^-.3093t))=ln(.03)
-.03093t=ln(.03)
t= (ln(.03)) / (-.3093t)
ln e^x = x
2007-04-10 13:12:26 · answer #1 · answered by ytrewq 3 · 0⤊ 0⤋
Loge(.03)=-.3093t
Logarithm in base e is called "Ln" instead of Log
Then Ln(.03)=-3.506
Then -.3093t=-3.506
divide -3.506 by -.3093 and find that
t=11.227
2007-04-10 13:14:20 · answer #2 · answered by Juan 3 · 0⤊ 0⤋
ln(.03)/-.3093 = t
2007-04-10 13:17:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋
ln (e^-.3093t) = ln (0.03)
The ln and e cancel out
-.3093t = ln (0.03)
Divide by -.3093
t = -ln(0.03) / .3093
2007-04-10 13:14:04 · answer #4 · answered by Blondie 3 · 0⤊ 0⤋
t=ln(.03)/(-.3093)
2007-04-10 13:10:55 · answer #5 · answered by bruinfan 7 · 0⤊ 0⤋