Question

2 log (x-1) = logs(4x + 1)

Answer 1

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)

Answer 2

Ответ: 2 log (x-1) = logs(4x + 1)

x=6

Объяснение: