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Question-145108




Question Number 145108 by imjagoll last updated on 02/Jul/21
Answered by liberty last updated on 02/Jul/21
 (( ((2^(x+1) /(32^(2x−2) )))^(1/(1/5)) ))^(1/(1/4)) = 64  ⇒ (( ((2^(x+1) /2^(10x−10) ))^5 ))^(1/(1/4)) = 2^6   ⇒(2^(−9x+11) )^(20)  = 2^6   ⇒(11−9x)=((10)/3)  ⇒9x = ((23)/3) ; x=((23)/(27)).
$$\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\sqrt[{\mathrm{1}/\mathrm{5}}]{\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{32}^{\mathrm{2}{x}−\mathrm{2}} }}}=\:\mathrm{64} \\ $$$$\Rightarrow\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\left(\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{2}^{\mathrm{10}{x}−\mathrm{10}} }\right)^{\mathrm{5}} }=\:\mathrm{2}^{\mathrm{6}} \\ $$$$\Rightarrow\left(\mathrm{2}^{−\mathrm{9}{x}+\mathrm{11}} \right)^{\mathrm{20}} \:=\:\mathrm{2}^{\mathrm{6}} \\ $$$$\Rightarrow\left(\mathrm{11}−\mathrm{9}{x}\right)=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{9}{x}\:=\:\frac{\mathrm{23}}{\mathrm{3}}\:;\:{x}=\frac{\mathrm{23}}{\mathrm{27}}.\: \\ $$

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