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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 −(3/4)x) ≤ ((√8))^x^3
64x2−34x⩽(8)x3
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)) :})) :}
64x2−34x⩽(8)x3(26)x2−34x⩽(232)x36x2−92x⩽32x312x2−9x⩽3x33x3−12x2+9x⩾0x(x2−4x+3)⩾0{x⩾0∧x2−4x+3⩾0∨x⩽0∧x2−4x+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
Commented by bemath last updated on 24/Sep/20
gave kudos sir
gavekudossir
Answered by Olaf last updated on 24/Sep/20
x(x−1)(x−3) ≥ 0 (1) ⇔ x∈[0;1]∪[3;+∞[ Then (1) : x(x^2 −4x+3) ≥ 0 x^3 −4x^2 +3x ≥ 0 4x^2 −3x ≤ x^3 (3/2)(4x^2 −3x) ≤ (3/2)x^3 6(x^2 −(3/4)x) ≤ (3/2)x^3 6ln2(x^2 −(3/4)x) ≤ ((3/2)ln2)x^3 (x^2 −(3/4)x)ln64 ≤ x^3 ln(√8) ln64^(x^2 −(3/4)x) ≤ ln((√8))^x^3 64^(x^2 −(3/4)x) ≤( (√8))^x^3 S = [0;1]∪[3;+∞[
x(x−1)(x−3)⩾0(1)⇔x∈[0;1]∪[3;+∞[Then(1):x(x2−4x+3)⩾0x3−4x2+3x⩾04x2−3x⩽x332(4x2−3x)⩽32x36(x2−34x)⩽32x36ln2(x2−34x)⩽(32ln2)x3(x2−34x)ln64⩽x3ln8ln64x2−34x⩽ln(8)x364x2−34x⩽(8)x3S=[0;1]∪[3;+∞[

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