Solve inequality in respect to monotonicity of logarithm

l0og (base sin(pie/6)) (7x-6) <= 3

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You need to replace `1/2` for `sin(pi/6)` to the base of logarithm, such that:

`log_(sin(pi/6)) (7x - 6) <= 3`

`log_(1/2) (7x - 6) <= 3`

By definition, if `f(x) = log_a x` and `a<1` yields that `f(x)` decreases.

Reasoning by analogy, yields:

`log_(1/2) (7x - 6) <= 3 => (7x - 6) >= (1/2)^3`

Adding 6 both sides, yields:

`7x >= 1/8 + 6 => 7x >= 49/8`

Dividing by 7 yields:

`x >= 7/8`

**Hence, evaluating the solution to the given inequality, using the properties of logarithmic function, yields that for all `x >= 7/8` , the inequality holds.**

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