1. e^ln3x=11
2. e^x+4/ e^4=1
3. (e^4)3x=e^4e^3x
4. e(2lnx-ln(x^2-3x+1)=1
5. e^2x-2 * e(x+2) + e^4 = 0
2007-08-18 18:05:15 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'll give you a few hints:
1. remember e^lny = y
3. Use your laws of exponents to simplify each side. Then remember from Alg II that if the bases are the same, then the exponents are equal
5. Again, use your laws of exponents to simplify this first, then solve.
2007-08-18 18:11:36 · answer #1 · answered by douglas 2 · 0⤊ 0⤋
1. 3x = 11 => x = 11/3
2. x+4-4 = 0 => x = 0
3. 3x = e^3x => numerical solution
4. x^2-3x+1 = x^2 => x = 1/3
5. no real solution
2007-08-19 01:26:43 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
I'd give it a shot, but I just can't be sure of your expressions. Take #2. Is it
[e^(x+4)] / e^4 = 1 or is it
(e^x + 4) / e^4 = 1 or is it
e^x + (4/e^4) = 1 ?
by strict rules, what you wrote means the 3rd one, but I'd bet you meant the 1st one.
2007-08-19 01:15:40 · answer #3 · answered by Philo 7 · 0⤊ 0⤋