I took ln of both sides so (3x+1)ln2=(x-2)ln3 but that didnt work, why not?
2006-06-22 18:33:51 · 6 answers · asked by znnyz 2 in Science & Mathematics ➔ Mathematics
It does work.
(3x+1)ln2=(x-2)ln3
Distribute ln2 on the left side and ln3 on the right side.
3x*ln2 + ln2 = x*ln3 - 2ln3
Subtract x*ln3 from each side
3x*ln2 - x*ln3 + ln2 = -2ln3
Subtract ln2 from each side
3x*ln2 - x*ln3 = -2ln3 - ln2
Factor an x from the left side
x(3ln2 - ln3) = -2ln3 - ln2
Divide both sides by 3ln2 - ln3
x = (-2ln3 -ln2)/(3ln2 -ln3)
x = -2.9469
2006-06-22 18:55:24 · answer #1 · answered by MsMath 7 · 2⤊ 0⤋
Actually, that will work. Since divide both sides by ln2: 3x+1=(x-2)(ln3/ln2) = (ln3/ln2)x - 2(ln3/ln2)
Isolate x:
(3-ln3/ln2)x = -1 - 2(ln3/ln2)
x = -2.946865368
2006-06-23 13:36:10 · answer #2 · answered by vishalarul 2 · 0⤊ 0⤋
2^(3x + 1) = 3^(x - 2)
(3x + 1)ln2 = (x - 2)ln3
3x(ln2) + ln2 = x(ln3) - 2ln3
3x(ln2) - x(ln3) = -ln2 - 2ln3
x(3ln2 - ln3) = -ln2 - 2ln3
x = (-ln2 - 2ln3)/(3ln2 - ln3)
x = -2.9469
2006-06-23 13:30:52 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
42
2006-06-23 01:36:27 · answer #4 · answered by Anonymous · 0⤊ 0⤋
2^(3x + 1)=3^(x - 2)
get (x - 2)th root of both sides
2^[(3x + 1)/(x - 2)] = 3
rewrite 3 into 2^log_2 3
2^[(3x + 1)/(x - 2)] = 2^log_2 3
Equate the exponents of equal bases
(3x + 1)/(x - 2) = log_2 3
Multiply x - 2
3x + 1 = (log_2 3) x - 2log_2 3
Transpose
3x - (log_2 3) x = -2log_2 3 - 1
factor
x(3 - log_2 3) = -2log_2 3 - 1
divide
x = (-2log_2 3 - 1)/(3 - log_2 3)
multiply -1/-1
x = (2 log_2 3 + 1)/(log_2 3 - 3)
or
x = -1.4150
^_^
2006-06-23 06:39:26 · answer #5 · answered by kevin! 5 · 0⤊ 0⤋
Mathgirl seems to be right. But we can also use common log can't we ?
2006-06-23 02:34:59 · answer #6 · answered by nayanmange 4 · 0⤊ 0⤋