Asked by Anonymous
Solve:
2log_3(x) - log_3(x-2) = 2
My Work:
log_3(x^2) - log_3(x-2) = log_3(3^2)
log_3 [(x^2) / (x-2)] = log_3(27)
(x^2) / (x-2) = 27
x^2 = (x-2)(27)
x^2 = 27x - 54
x^2 - 27x = -54
27x = -54
x = -2
Textbook Answer: 3 or 6
Where did I go wrong and how should I fix it?
2log_3(x) - log_3(x-2) = 2
My Work:
log_3(x^2) - log_3(x-2) = log_3(3^2)
log_3 [(x^2) / (x-2)] = log_3(27)
(x^2) / (x-2) = 27
x^2 = (x-2)(27)
x^2 = 27x - 54
x^2 - 27x = -54
27x = -54
x = -2
Textbook Answer: 3 or 6
Where did I go wrong and how should I fix it?
Answers
Answered by
Damon
log_3(x^2/(x-2)) = 2 nothing changed on the right
x^2/(x-2) = 3^2 = 9
x^2 = 9x - 18
x^2 -9x + 18 = 0
(x-3)(x-6) = 0
for once I agree with your book, 3 or 6
x^2/(x-2) = 3^2 = 9
x^2 = 9x - 18
x^2 -9x + 18 = 0
(x-3)(x-6) = 0
for once I agree with your book, 3 or 6
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