ln (x-1) + ln (x-5) = ln(3x+7)
2006-11-27 02:23:33 · 5 answers · asked by Dominique W 1 in Science & Mathematics ➔ Mathematics
ln (x-1) + ln (x-5) = ln(3x+7)
ln( (x-1)(x-5) ) = ln(3x+7)
so
(x-1)(x-5) = (3x+7)
x^2 - 6x + 5 = 3x + 7
x^2 -9x -2 =0
x= (9+- sqrt( 81 -4(-2)) )/2
x = ( 9 +- sqrt(89) )/2
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2006-11-27 02:37:37 · answer #1 · answered by Anonymous · 2⤊ 0⤋
ln(x - 1) + ln(x - 5) = ln(x - 1)(x - 5) = ln(3x + 7)
so
(x - 1)(x - 5) = x^2 - 6x + 5 = 3x + 7 and
x^2 - 9x - 2 = 0
the roots are
x = [9 +- (89)^(1/2)] / 2
The negative root is not a solution of the original equation, so the unique solution is
x = [9 + (89)^(1/2)] / 2
2006-11-27 10:31:25 · answer #2 · answered by airtime 3 · 0⤊ 0⤋
ln (x-1) + ln (x-5) = ln(3x+7)
ln( (x-1)(x-5) ) = ln(3x+7)
=>
(x-1)(x-5) = (3x+7)
x^2 - 6x + 5 = 3x + 7
x^2 -9x -2 =0
x= (9+- sqrt( 81 -4(-2)) )/2
x = ( 9 +- sqrt(89) )/2
.
2006-11-29 13:03:05 · answer #3 · answered by lobis3 5 · 0⤊ 0⤋
ln(x-1) +ln(x-5) =ln(3x+7)
ln[(x-1)(x-5)] = ln(3x+7)
e^ln(x-1)(x-5) = e^ln(3x+7)
(x-1)(x-5) = 3x+7
x^2 -6x + 5 = 3x + 7
x^2-9x - 2 = 0
x= [9 +or - sqrt(81-4(1)(-2)]/2
x = 9/2 + or - 0.5*sqrt(89)
x = 9/2 + 0.5*sqrt(89) {reject other solution}
2006-11-27 10:51:22 · answer #4 · answered by ironduke8159 7 · 1⤊ 0⤋
log(x-1)(x-5)=log(3x+7)
removing the log
(x^2-6x+5)=3x+7
adding -3x-7
x^2-9x-2=0
x=[2+/-rt(81+8)]/2
=[2+/-rt89]/2
2006-11-27 10:34:21 · answer #5 · answered by raj 7 · 0⤊ 2⤋