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Question Number 181779 by DAVONG last updated on 30/Nov/22

3^(4x−11) +3^(10−x^2 ) +3^((x−2)^2 ) =6x−x^2

$$\mathrm{3}^{\mathrm{4x}−\mathrm{11}} +\mathrm{3}^{\mathrm{10}−\mathrm{x}^{\mathrm{2}} } +\mathrm{3}^{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} } =\mathrm{6x}−\mathrm{x}^{\mathrm{2}} \\ $$

Answered by manxsol last updated on 01/Dec/22

6x−x^2 ≫0⇒0≪x≪6  −x^2 +6x−t=0⇒△=6^2 −4t  △≫0⇒t≪9  3^(4x−11) ≪3^1 ⇒x≪3  3^(10−x^2 ) ≪3^1 ⇒−3≪x∪x≫3  3^((x−2)^2 ≪) 3^(1⇒)  1≪x≪3  t≪9  solution x=3

$$\mathrm{6}{x}−{x}^{\mathrm{2}} \gg\mathrm{0}\Rightarrow\mathrm{0}\ll{x}\ll\mathrm{6} \\ $$$$−{x}^{\mathrm{2}} +\mathrm{6}{x}−{t}=\mathrm{0}\Rightarrow\bigtriangleup=\mathrm{6}^{\mathrm{2}} −\mathrm{4}{t} \\ $$$$\bigtriangleup\gg\mathrm{0}\Rightarrow{t}\ll\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{4}{x}−\mathrm{11}} \ll\mathrm{3}^{\mathrm{1}} \Rightarrow{x}\ll\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{10}−{x}^{\mathrm{2}} } \ll\mathrm{3}^{\mathrm{1}} \Rightarrow−\mathrm{3}\ll{x}\cup{x}\gg\mathrm{3} \\ $$$$\mathrm{3}^{\left({x}−\mathrm{2}\right)^{\mathrm{2}} \ll} \mathrm{3}^{\mathrm{1}\Rightarrow} \:\mathrm{1}\ll{x}\ll\mathrm{3} \\ $$$${t}\ll\mathrm{9}\:\:{solution}\:{x}=\mathrm{3} \\ $$

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