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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

$$\sqrt{\mathrm{4}^{{x}} −\mathrm{5}.\mathrm{2}^{{x}+\mathrm{1}} +\mathrm{25}}\:+\sqrt{\mathrm{9}^{{x}} −\mathrm{2}.\mathrm{3}^{{x}+\mathrm{2}} +\mathrm{17}}\:\leqslant\:\mathrm{2}^{{x}} −\mathrm{5} \\ $$

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)

$$\sqrt{\mathrm{4}^{{x}} −\mathrm{5}.\mathrm{2}^{{x}+\mathrm{1}} +\mathrm{25}}\:+\sqrt{\mathrm{9}^{{x}} −\mathrm{2}.\mathrm{3}^{{x}+\mathrm{2}} +\mathrm{17}}\:\leqslant\:\mathrm{2}^{{x}} −\mathrm{5} \\ $$$${note}\:{that}\:\mathrm{2}^{{x}} −\mathrm{5}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\sqrt{\left(\mathrm{2}^{{x}} −\mathrm{5}\right)^{\mathrm{2}} }+\sqrt{\mathrm{9}^{{x}} −\mathrm{18}.\mathrm{3}^{{x}} +\mathrm{17}}\:\leqslant\mathrm{2}^{{x}} −\mathrm{5} \\ $$$$\mid\mathrm{2}^{{x}} −\mathrm{5}\mid\:+\sqrt{\mathrm{9}^{{x}} −\mathrm{18}.\mathrm{3}^{{x}} +\mathrm{17}}\:\leqslant\mathrm{2}^{{x}} −\mathrm{5} \\ $$$$\sqrt{\mathrm{9}^{{x}} −\mathrm{18}.\mathrm{3}^{{x}} +\mathrm{17}}\:\leqslant\:\mathrm{0} \\ $$$${let}\:\mathrm{3}^{{x}} \:=\:{t}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{18}.{t}+\mathrm{17}\:\leqslant\:\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}−\mathrm{17}\right)\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:{t}=\mathrm{1}\:\vee{t}=\mathrm{17} \\ $$$$\Rightarrow\mathrm{3}^{{x}} =\mathrm{1}\Rightarrow{x}=\mathrm{0}\:\left({rejected}\right) \\ $$$$\Rightarrow\mathrm{3}^{{x}} \:=\:\mathrm{17}\:\Rightarrow{x}\:=\:^{\mathrm{3}} \mathrm{log}\:\mathrm{17}\:\left({acceptable}\right) \\ $$$$ \\ $$

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