Solve the equation.
17-4e^-3t = 8
I ended up getting a non-real answer. I ended up trying to take the natural log of -25/4 and I get a non-real answer in my calculator. Did i do something wrong?
2007-11-06 08:33:56 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If you subtract 17 from both sides you get -9 on the right. Divide both sides by -4 to get 9/4 on the right. This is what you should be taking the natural log of.
2007-11-06 08:38:39 · answer #1 · answered by Demiurge42 7 · 0⤊ 0⤋
Natural log is defined only on the interval (0, â).
17-4e^-3t = 8
9 = 4e^-3t
9/4 = e^-3t
ln(9/4) = -3t
[ln(9/4)]/(-3) = t
-ln(9/4)/3 = t
2007-11-06 16:38:00 · answer #2 · answered by richarduie 6 · 0⤊ 0⤋
First 17-4e^3t = 8
becomes -4e^3t = -9
divide by -4
gives you e^3t = 2.25
Take Ln or natural log of both sides gives you
Ln(3^3t) = Ln(2.25) => 3t = .8109302...
Divide by 3
T= .2703100
Checked with calulator, correct.
2007-11-06 16:47:38 · answer #3 · answered by honest guy 4 · 0⤊ 0⤋
4e^-3t = 17-8 = 9
e^-3t = 9/4
Take ln,
-3t = ln(9/4)
t = -(1/3)ln(9/4)
2007-11-06 16:37:57 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋
17-4e^-3t = 8
-4e^(-3t) = 8 -17 = -9
e^(-3t) = 9/4
-3t = ln(9/4)
t = (-1/3)(ln(9/4)) = -0.2703
2007-11-06 16:37:40 · answer #5 · answered by Any day 6 · 2⤊ 0⤋