Show work please
A.5
B.3 or -2
C.3
D.1
Solve for x: 2 log3(x)=log3(x+6)?
2009-04-30 4:18 pm
回答 (5)
2009-04-30 4:38 pm
✔ 最佳答案
2 log[3] x = log[3] (x+6)log[3] x² = log[3] (x+6)
Compare: (cancel off both side's log[3])
x² = x + 6
x² - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2, 3
Since 2 log[3] (-2) does not exist, x = 3
Answer: C
Done
2009-04-30 11:27 pm
2 log3(x)=log3(x+6)
log3 (x^2) = lof3(x+6)
x^2=x+6
x^2-x-6=0
x^2-3x+2x-6=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=3, x=-2
x=3
log3 (x^2) = lof3(x+6)
x^2=x+6
x^2-x-6=0
x^2-3x+2x-6=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=3, x=-2
x=3
2009-04-30 11:25 pm
2log_3(x) = log_3(x + 6)
log_3(x^2) = log_3(x + 6)
x^2 = x + 6
x^2 - x - 6 = 0
x^2 + 2x - 3x - 6 = 0
(x^2 + 2x) - (3x + 6) = 0
x(x + 2) - 3(x + 2) = 0
(x + 2)(x - 3) = 0
x + 2 = 0
x = -2
x - 3 = 0
x = 3
â´ x = -2 , 3
(answer B)
log_3(x^2) = log_3(x + 6)
x^2 = x + 6
x^2 - x - 6 = 0
x^2 + 2x - 3x - 6 = 0
(x^2 + 2x) - (3x + 6) = 0
x(x + 2) - 3(x + 2) = 0
(x + 2)(x - 3) = 0
x + 2 = 0
x = -2
x - 3 = 0
x = 3
â´ x = -2 , 3
(answer B)
2009-04-30 11:24 pm
2 log3(x)=log3(x+6)
log3(x^2)=log3(x+6)
x^2=x+6
x^2-x-6=0
(x-3)(x+2)
x=3 or x=-2
x= -2 not allowed
Then x=3
log3(x^2)=log3(x+6)
x^2=x+6
x^2-x-6=0
(x-3)(x+2)
x=3 or x=-2
x= -2 not allowed
Then x=3
2009-04-30 11:23 pm
2 log3(x)=log3(x+6)
log3(x^2)=log3(x+6)
x^2=x+6
x^2-x-6=0
(x-3)(x+2)
x=3 or x=-2
x= -2 not allowed
Then x=3
Answer it's C.
log3(x^2)=log3(x+6)
x^2=x+6
x^2-x-6=0
(x-3)(x+2)
x=3 or x=-2
x= -2 not allowed
Then x=3
Answer it's C.
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