2007-09-15 02:41:56 · 3 answers · asked by Shinichi 1 in Science & Mathematics ➔ Mathematics
Sorry, the solution set of this equation is empty.
Proof:
log _5(x-7)/(x-3) = 1
(x-7)/(x-3) = 5
x-7 = 5(x-3) = 5x-15
4x = 8
x = 2
but
log(2-7) doesn't exist since the log of
a negative number is not real.
So there is no solution to this problem.
2007-09-15 03:25:00 · answer #1 · answered by steiner1745 7 · 0⤊ 0⤋
log (x-7) to the base 5 minus the log of (x-3) to the base 5 = 1
=> log (x-7) - log (x-3) =1
=> log[ (x-7)/ (x-3) ] = 1
=> (x-7)/ (x-3) =5^1
=> x-7 = 5( x-3 )
the rest is easy
2007-09-15 02:49:41 · answer #2 · answered by harry m 6 · 0⤊ 0⤋
log5 (x - 7) - log5 (x - 3) = 1
log5 [(x - 7)/(x - 3)] = 1
(x - 7)/(x - 3) = 5^1
x - 7 = 5(x - 3)
x - 7 = 5x - 15
8 = 4x
x = 2
x = 2 rejected for logarithm to be defined. Therefore, there is no solution set to the equation.
2007-09-15 03:47:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋