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Question-139746




Question Number 139746 by mathlove last updated on 01/May/21
Answered by bramlexs22 last updated on 01/May/21
log _2 (x)=t   ⇒t+(1/t) = e+e^(−1)   ⇒t^2 −(e+e^(−1) )t+1=0  ⇒t = ((e+e^(−1)  ± (√((e+e^(−1) )^2 −4)))/2)  ⇒t = ((e+e^(−1)  ± (√(e^2 −2+e^(−2) )))/2)  ⇒ log _2 (x) = ((e+e^(−1) ± (e−e^(−1) ))/2)  (1) log _2 (x)=((e+e^(−1) +e−e^(−1) )/2)   ⇒x = 2^e   (2) log _2 (x)=((e+e^(−1) −e+e^(−1) )/2)   ⇒ x = 2^e^(−1)   = 2^(1/e)
$$\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{t} \\ $$$$\:\Rightarrow\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\:=\:\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} −\left(\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \right)\mathrm{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}\:=\:\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \:\pm\:\sqrt{\left(\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}\:=\:\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \:\pm\:\sqrt{\mathrm{e}^{\mathrm{2}} −\mathrm{2}+\mathrm{e}^{−\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \pm\:\left(\mathrm{e}−\mathrm{e}^{−\mathrm{1}} \right)}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\frac{\mathrm{e}+\cancel{\mathrm{e}^{−\mathrm{1}} }+\mathrm{e}−\cancel{\mathrm{e}^{−\mathrm{1}} }}{\mathrm{2}}\: \\ $$$$\Rightarrow\mathrm{x}\:=\:\mathrm{2}^{\mathrm{e}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\frac{\cancel{\mathrm{e}}+\mathrm{e}^{−\mathrm{1}} −\cancel{\mathrm{e}}+\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\: \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\mathrm{2}^{\mathrm{e}^{−\mathrm{1}} } \:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{e}}} \\ $$$$ \\ $$

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