64^(x^2 −(3/4)x) ≤ ((√8))^x^3


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Question Number 115238 by bemath last updated on 24/Sep/20

$64^{x^{2} - \frac{3}{4} x} ⩽ {(\sqrt{8})}^{x^{3}}$
	Answered by Rasheed.Sindhi last updated on 24/Sep/20
	$64^(x^2 −(3/4)x) ≤ ((√8))^x^3 (2^6 )^(x^2 −(3/4)x) ≤ (2^(3/2) )^x^3 6x^2 −(9/2)x≤(3/2)x^3 12x^2 −9x≤3x^3 3x^3 −12x^2 +9x≥0 x(x^2 −4x+3)≥0 { ((x≥0 ∧ x^2 −4x+3≥0)),(( ∨)),((x≤0 ∧ x^2 −4x+3≤0)) :} { ((x≥0 ∧{(x−1)(x−3)}≥0)),(( ∨)),((x≤0 ∧{(x−1)(x−3) }≤0)) :} { ((x≥0 ∧ { ((x≥1 ∧ x≥3)),(( ∨)),((x≤1 ∧ x≤3)) :})),(( ∨)),((x≤0 ∧ { ((x≤1 ∧ x≥3 )),(( ∨)),((x≥1 ∧ x≤3)) :})) :}$
	$64^{x^{2} - \frac{3}{4} x} ⩽ {(\sqrt{8})}^{x^{3}}$ ${(2^{6})}^{x^{2} - \frac{3}{4} x} ⩽ {(2^{\frac{3}{2}})}^{x^{3}}$ $6 x^{2} - \frac{9}{2} x ⩽ \frac{3}{2} x^{3}$ $12 x^{2} - 9 x ⩽ 3 x^{3}$ $3 x^{3} - 12 x^{2} + 9 x ⩾ 0$ $x (x^{2} - 4 x + 3) ⩾ 0$ ${\begin{cases} x ⩾ 0 \land x^{2} - 4 x + 3 ⩾ 0 \\ \lor \\ x ⩽ 0 \land x^{2} - 4 x + 3 ⩽ 0 \end{cases}$ ${\begin{cases} x ⩾ 0 \land {(x - 1) (x - 3)} ⩾ 0 \\ \lor \\ x ⩽ 0 \land {(x - 1) (x - 3)} ⩽ 0 \end{cases}$ ${\begin{cases} x ⩾ 0 \land {\begin{cases} x ⩾ 1 \land x ⩾ 3 \\ \lor \\ x ⩽ 1 \land x ⩽ 3 \end{cases} \\ \lor \\ x ⩽ 0 \land {\begin{cases} x ⩽ 1 \land x ⩾ 3 \\ \lor \\ x ⩾ 1 \land x ⩽ 3 \end{cases} \end{cases}$
	Commented by bemath last updated on 24/Sep/20

	$g a v e k u d o s s i r$
	Answered by Olaf last updated on 24/Sep/20

	$x (x - 1) (x - 3) ⩾ 0 (1)$ $\Leftrightarrow x \in [0; 1] \cup [3; + \infty [$ $Then (1) : x (x^{2} - 4 x + 3) ⩾ 0$ $x^{3} - 4 x^{2} + 3 x ⩾ 0$ $4 x^{2} - 3 x ⩽ x^{3}$ $\frac{3}{2} (4 x^{2} - 3 x) ⩽ \frac{3}{2} x^{3}$ $6 (x^{2} - \frac{3}{4} x) ⩽ \frac{3}{2} x^{3}$ $6 \ln 2 (x^{2} - \frac{3}{4} x) ⩽ (\frac{3}{2} \ln 2) x^{3}$ $(x^{2} - \frac{3}{4} x) \ln 64 ⩽ x^{3} \ln \sqrt{8}$ ${\ln 64}^{x^{2} - \frac{3}{4} x} ⩽ \ln {(\sqrt{8})}^{x^{3}}$ $64^{x^{2} - \frac{3}{4} x} ⩽ {(\sqrt{8})}^{x^{3}}$ $S = [0; 1] \cup [3; + \infty [$
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