2007-12-26 18:04:53 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
By inspection:
log(x+6) is greater than log(x+2)
Therefore:
log(x+2) - log(x+6) must be negative.
No solution exists.
2007-12-26 18:31:08 · answer #1 · answered by Valithor 4 · 2⤊ 1⤋
log (x + 2) - log (x + 6) = 2
Using the quotient law:
log [(x + 2) / (x + 6)] = 2
Write in exponential form:
10^2 = (x + 2) / (x + 6)
100(x + 6) = x + 2
100x + 600 = x + 2
99x = -598
x = approx. -6.04
It appears that there is no solution to this equation, as x can not make the argument less than or equal to 0 in either of the logarithms.
2007-12-26 18:11:15 · answer #2 · answered by Jacob A 5 · 2⤊ 1⤋
log(x + 2) - log(x + 6) = 2
log(x + 2) / (x + 6) = 2
10^2 = (x + 2) / (x + 6)
100*(x + 6) = (x + 2)
100x + 600 = x + 2
99x = -598
x = -598/99 or - 6,04040404
Therefore, as any number "y" in which 10^y = negative result, there is no solution, S = { }
FCH
2007-12-26 18:09:47 · answer #3 · answered by Dr. Milk 4 · 2⤊ 2⤋
log(x + 2) - log(x + 6) = 2
log[ (x + 2) / (x + 6) ] = 2
(x + 2) / (x + 6) = 10^2
x + 2 = 100x + 600
99x + 598 = 0
x = -598 / 99
x = -6.0404040404 ... :-o!
But the negative real numbers are not part of the domain of the log function because it is all real numbers ruling the zero out.
2007-12-26 19:35:37 · answer #4 · answered by fraukka 3 · 1⤊ 2⤋
log [ (x + 2) / (x + 6) ] = 2
[ (x + 2) / (x + 6) ] = 10²
(x + 2) / (x + 6) = 100
x + 2 = 100x + 600
99x = - 598
x = - 598 / 99
2007-12-26 21:09:46 · answer #5 · answered by Como 7 · 3⤊ 1⤋