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Question Number 115268 by bobhans last updated on 24/Sep/20

(√(4^x −5.2^(x+1) +25)) +(√(9^x −2.3^(x+2) +17)) ≤ 2^x −5

4x−5.2x+1+25+9x−2.3x+2+17⩽2x−5

Answered by john santu last updated on 24/Sep/20

(√(4^x −5.2^(x+1) +25)) +(√(9^x −2.3^(x+2) +17)) ≤ 2^x −5 note that 2^x −5 ≥ 0 ⇒(√((2^x −5)^2 ))+(√(9^x −18.3^x +17)) ≤2^x −5 ∣2^x −5∣ +(√(9^x −18.3^x +17)) ≤2^x −5 (√(9^x −18.3^x +17)) ≤ 0 let 3^x = t ⇒t^2 −18.t+17 ≤ 0 (t−1)(t−17)≤ 0 ⇒ t=1 ∨t=17 ⇒3^x =1⇒x=0 (rejected) ⇒3^x = 17 ⇒x =^3 log 17 (acceptable)

4x−5.2x+1+25+9x−2.3x+2+17⩽2x−5notethat2x−5⩾0⇒(2x−5)2+9x−18.3x+17⩽2x−5∣2x−5∣+9x−18.3x+17⩽2x−59x−18.3x+17⩽0let3x=t⇒t2−18.t+17⩽0(t−1)(t−17)⩽0⇒t=1∨t=17⇒3x=1⇒x=0(rejected)⇒3x=17⇒x=3log17(acceptable)

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