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Question Number 106695 by bobhans last updated on 06/Aug/20

    #bobhans#  3^x −2^(x+1)  ≤ (√(2.9^x −10.6^x +2^(2x+3) ))  find the solution set

You can't use 'macro parameter character #' in math mode3x2x+12.9x10.6x+22x+3findthesolutionset

Answered by bemath last updated on 06/Aug/20

        ^(@bemath@)   ((3^x −2.2^x )/2^x ) ≤ ((√(2.9^x −10.6^x +8.(2^x )^2 ))/2^x )  ((3/2))^x −2 ≤ (√(2((3/2))^(2x) −10((3/2))^x +8))  set ((3/2))^x = t   t−2 ≤ (√(2(t−1)(t−4)))  defined on (t−1)(t−4)≥0  ⇒ t ≤1 ∪ t ≥ 4...(1)  squaring both sides  t^2 −4t+4 ≤ 2t^2 −10t+8  −t^2 +6t−4 ≤ 0 ; t^2 −6t+4 ≥ 0  (t−3)^2 −5≥0  t≤3−(√5) ∪t ≥ 3+(√5) ...(2)  from (1)∩(2) we get → { ((t ≤1)),((t≥3+(√5))) :}   { ((((3/2))^x ≤1 ⇒ x ≤ 0)),((((3/2))^x ≥ 3+(√5) ⇒x ≥log _(((3/2))) (3+(√5)) )) :}  the solution set is x∈(−∞,0 ] ∪  [ log _(((3/2))) (3+(√5)) ,∞ )

@bemath@3x2.2x2x2.9x10.6x+8.(2x)22x(32)x22(32)2x10(32)x+8set(32)x=tt22(t1)(t4)definedon(t1)(t4)0t1t4...(1)squaringbothsidest24t+42t210t+8t2+6t40;t26t+40(t3)250t35t3+5...(2)from(1)(2)weget{t1t3+5{(32)x1x0(32)x3+5xlog(32)(3+5)thesolutionsetisx(,0][log(32)(3+5),)

Commented by bobhans last updated on 06/Aug/20

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