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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