I'm stuck with an exponent question in my chem HW that I don't know how to solve. Can anyone offer some help? I'll give a simplified example of the type of problem:
.2 + e^(4/t) = 5*e^(8/t). I'm trying to solve for t. THanks to anyone who can help!!!
2007-11-11 14:58:25 · 4 answers · asked by Michael T 2 in Science & Mathematics ➔ Mathematics
Let x = e^(4/t)
then 0.2 + x = 5x^2
5x^2 - x - 2 = 0
x = (1 ± √(1 + 4*5*0.2))/10
x = (1 ± √5)/10
x ≈ - 0.1236068, 0.3236068
e^(4/t) ≈ 0.3236068
(The negative solution is unusable.)
4/t ≈ ln(0.3236068)
t ≈ 4/ln(0.3236068)
t ≈ - 3.545389 ≈ - 3.55
2007-11-11 15:15:46 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
Let u = e^(4/t). u > 0
5u^2-u-0.2 = 0
Use quadratic formula to get u
t = 4/lnu
2007-11-11 23:04:06 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
5[e^(4/t)]^2 - e^(4/t) - .2 = 0
The roots are:
e^(4/t) = [-1 +/- sqrt(25 - .8)]/10
= [-1 +/- 4.919]/10
The answer must be +, so e^(4/t) = .3919
4/t = ln (.3919)
t= -4.27
2007-11-11 23:16:59 · answer #3 · answered by dooner75 3 · 0⤊ 0⤋
Just put it in your graphing calculator
solve (0.2 + e^(4/t)=5e^(8/t),t)
getting t = -3.54
2007-11-11 23:03:27 · answer #4 · answered by Tim B 3 · 0⤊ 0⤋