What is log_{0.5}(x^2-4x+3)<= 1 ?

The solution to log_{0.5}(x^2-4x+3) = 3.22474…

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log_{0.5}(x^2-4x+3)<= 1

lo g_{0.5} (x^{2} - 4 x + 3) \leq 1

x \leq 0.77525 \dots or x \geq 3.22474 \dots

Interval Notation

(- \infty, 0.77525 \dots] \cup [3.22474 \dots, \infty)

Solution steps

lo g_{0.5} (x^{2} - 4 x + 3) \leq 1

Apply log rules

x^{2} - 4 x + 3 \geq 0.5

x^{2} - 4 x + 3 \geq 0.5 : x \leq 0.77525 \dots or x \geq 3.22474 \dots

x \leq 0.77525 \dots or x \geq 3.22474 \dots

lo g_{n} (n + 1) < lo g_{n - 1} (n)

lo g_{0.2} (x + 5) > 0

\frac{x}{( lo g _{\frac{1}{2}} ( 2 x + 3 ) )} > 0

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

lo g_{10} (2 - x) > lo g_{10} (1 + 2 x)

What is log_{0.5}(x^2-4x+3)<= 1 ?
The solution to log_{0.5}(x^2-4x+3)<= 1 is x<= 0.77525…\lor x>= 3.22474…