x-2-4x-3-x-2-6x-4-1-

Question Number 130337 by EDWIN88 last updated on 24/Jan/21

{(x^{2} - 4 x + 3)}^{x^{2} - 6 x + 4} ⩽ 1

Answered by mathmax by abdo last updated on 24/Jan/21

e \Rightarrow e^{(x^{2} - 6 x + 4) \ln (x^{2} - 4 x + 3)} ⩽ 1

x^{2} - 4 x + 3 = 0 \to Δ^{^{'}} = 4 - 3 = 1 \Rightarrow x_{1} = \frac{2 + 1}{1} = 3 and x_{2} = \frac{2 - 1}{1} = 1

the equation is dfined on] - \infty, 1 [\cup] 3, + \infty [

e \Rightarrow (x^{2} - 6 x + 4) \ln (x^{2} - 4 x + 3) ⩽ 0

x^{2} - 6 x + 4 = 0 \to Δ^{^{'}} = 9 - 4 = 5 \Rightarrow x_{1} = 3 + \sqrt{5} and x_{2} = 3 - \sqrt{5}

\ln (x^{2} - 4 x + 3) > 0 \Rightarrow x^{2} - 4 x + 3 > 1 \Rightarrow x^{2} - 4 x + 2 > 0 \Rightarrow

x^{2} - 4 x + 4 - 2 > 0 \Rightarrow {(x - 2)}^{2} - 2 > 0 \Rightarrow (x - 2 - \sqrt{2}) (x - 2 + \sqrt{2}) > 0 \Rightarrow

x \in] - \infty, 2 - \sqrt{2} [\cup] 2 + \sqrt{2}, + \infty [

\ln (x^{2} - 4 x + 3) < 0 \Rightarrow x^{2} - 4 x + 3 < 1 \Rightarrow x^{2} - 4 x + 2 < 0 \Rightarrow

] 2 - \sqrt{2}, 2 + \sqrt{2} [the problem lead to find sgn of (x^{2} - 6 x + 4) \ln (x^{2} - 4 x + 3)

\dots be continued \dots

Answered by MJS_new last updated on 24/Jan/21

p \in R \land q \in R \land r \in Z

p^{q} = 1

(1) p = 1

(2) q = 0 \land p \neq 0

(3) p = - 1 \land q = 2 r

(1) x^{2} - 4 x + 3 = 1 \Rightarrow x = 2 \pm \sqrt{2}

(2) x^{2} - 6 x + 4 = 0 \Rightarrow x = 3 \pm \sqrt{5}

x^{2} - 4 x + 3 \neq 0 in both cases

(3) x^{2} - 4 x + 3 = - 1 \Rightarrow x = 2

x^{2} - 6 x + 4 = - 4 in this case

both p and q are parabolas \Rightarrow

solution for p^{q} ⩽ 1 is

2 - \sqrt{2} ⩽ x ⩽ 3 - \sqrt{5} \lor x = 2 \lor 2 + \sqrt{2} ⩽ x ⩽ 3 + \sqrt{5}

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