What Values of "x" cannot possibly be solutions of the equation loga (3x + 1) = 2
******The loga is actually log with the subscript "a". Log with "a" underneath it. I wasen't sure how to represent that on the keyboard******
2006-11-20 12:33:33 · 6 answers · asked by Shawn 1 in Science & Mathematics ➔ Mathematics
when x < -1/3, the equation do NOT have solution.
2006-11-20 12:37:47 · answer #1 · answered by shamu 2 · 0⤊ 0⤋
loga (3x + 1) = 2
(3x + 1) > 0
So x > -â
The value of a has nothing to do with this condition (except a >0)
This is of course assuming that we are solving it for a, x being real numbers otherwise there is a whole new set of conditions established once we enter the complex number field.
2006-11-20 20:40:02 · answer #2 · answered by Wal C 6 · 0⤊ 0⤋
Well, there is no log for negative numbers...so that means that x>-1/3 but more importantly:
the equation is equivalent to
(3x+1) > a^2,
3x > a^2-1
x > a^2/3 -1/3
2006-11-20 20:43:53 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
Use math, not the keyboard. loga (3x + 1)=ln(3x+1)/ln(a)
Ln(3x+1)=2 ln(a)
3x+1=a^2
3x=a^2-1
x=(a^2-1)/3
2006-11-20 20:44:50 · answer #4 · answered by Anonymous · 0⤊ 0⤋
the above are all wrong. x <= - 1/3
2006-11-20 21:23:05 · answer #5 · answered by Sora Aoi 2 · 0⤊ 0⤋
that means that a^2 = 3X +1
For x to be unsolvable, 3X +1 = 0
x=(-1)/3 = -0.333333.....
2006-11-20 20:37:48 · answer #6 · answered by rizwan_tz 1 · 0⤊ 0⤋