Asked by Anonymous
Log_8 = Log with the base of 8
log_8 (x-3) - log_8 (3x+11) = log_8 [1 / (x+1)]
I use the quotient law for the left side and eliminate log_8. And then what?
My Work:
log_8 [(x-3)/(3x+11)] = log_8 [1 / (x+1)]
[(x-3)/(3x+11)] = [1 / (x+1)]
Answer: 7
log_8 (x-3) - log_8 (3x+11) = log_8 [1 / (x+1)]
I use the quotient law for the left side and eliminate log_8. And then what?
My Work:
log_8 [(x-3)/(3x+11)] = log_8 [1 / (x+1)]
[(x-3)/(3x+11)] = [1 / (x+1)]
Answer: 7
Answers
Answered by
Damon
multiply both sides by (3x+11)(x+1)
(x-3)(x+1) = 3x+11
x^2 - 2x - 3 = 3x + 11
x^2 - 5x -14 = 0
(x-7)(x+2) = 0
ok?
(x-3)(x+1) = 3x+11
x^2 - 2x - 3 = 3x + 11
x^2 - 5x -14 = 0
(x-7)(x+2) = 0
ok?
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