What is log_{3}(x+1)+log_{3}(x+7)>= 3 ?

The solution to log_{3}(x+1)+log_{3}(x+7)>= 3 is x>= 2

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log_{3}(x+1)+log_{3}(x+7)>= 3

lo g_{3} (x + 1) + lo g_{3} (x + 7) \geq 3

x \geq 2

Interval Notation

[2, \infty)

Solution steps

lo g_{3} (x + 1) + lo g_{3} (x + 7) \geq 3

Apply log rules

(x + 1) (x + 7) \geq 27

(x + 1) (x + 7) \geq 27 : x \leq - 10 or x \geq 2

x \leq - 10 or x \geq 2

Find positive values for logs:

x > - 1

Combine the intervals

(x \leq - 10 or x \geq 2) and x > - 1

Merge Overlapping Intervals

x \geq 2

lo g_{x} (6 - x) > 2

lo g_{4} (- x) > 0

lo g_{5} (2 x - 1) < lo g_{5} (x + 7)

x \cdot ln (x^{3}) > 0

lo g_{2} (x + 1) < 2

What is log_{3}(x+1)+log_{3}(x+7)>= 3 ?
The solution to log_{3}(x+1)+log_{3}(x+7)>= 3 is x>= 2