Объяснение:
2log(x - 1) = log[(x - 1)^2]
Therefore, the equation becomes:
log[(x - 1)^2] = log(4x + 1)
Using the property that states that if log a = log b, then a = b, we can equate the expressions inside the logarithms:
(x - 1)^2 = 4x + 1
Expanding the left side and simplifying the equation, we get:
x^2 - 6x + 4 = 0
Using the quadratic formula, we can solve for x:
x = [6 ± sqrt(6^2 - 4(1)(4))]/(2)
x = [6 ± sqrt(32)]/2
x = 3 ± sqrt(8)
x = 3 + sqrt(8) or x = 3 - sqrt(8)
Ответ: 2 log (x-1) = logs(4x + 1)
x=6
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Answers & Comments
Объяснение:
2log(x - 1) = log[(x - 1)^2]
Therefore, the equation becomes:
log[(x - 1)^2] = log(4x + 1)
Using the property that states that if log a = log b, then a = b, we can equate the expressions inside the logarithms:
(x - 1)^2 = 4x + 1
Expanding the left side and simplifying the equation, we get:
x^2 - 6x + 4 = 0
Using the quadratic formula, we can solve for x:
x = [6 ± sqrt(6^2 - 4(1)(4))]/(2)
x = [6 ± sqrt(32)]/2
x = 3 ± sqrt(8)
x = 3 + sqrt(8) or x = 3 - sqrt(8)
Ответ: 2 log (x-1) = logs(4x + 1)
x=6
Объяснение: