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Question Number 153840 by liberty last updated on 11/Sep/21
    log _e (x)+log _x (e)+log _(((e/x))) (x)=(5/2)   x=?
loge(x)+logx(e)+log(ex)(x)=52x=?
Answered by Rasheed.Sindhi last updated on 11/Sep/21
 log _e (x)+log _x (e)+log _(((e/x))) (x)=(5/2)  ((log_e  x)/(log_e  e))+((log_e e)/(log_e x))+((log_e x)/(log_e ((e/x))))=(5/2)  log_e  x+(1/(log_e  x))+((log_e  x)/(log_e e−log_e  x))=(5/2)  log_e  x+(1/(log_e  x))+((log_e  x)/(1−log_e  x))=(5/2)  log_e  x=y (say)  y+(1/y)+(y/(1−y))=(5/2)  y(2y(1−y))+2(1−y)+2y^2 =5y(1−y)  2y^2 −2y^3 +2−2y+2y^2 −5y+5y^2 =0  2y^3 −9y^2 +7y−2=0  ....  ...
loge(x)+logx(e)+log(ex)(x)=52logexlogee+logeelogex+logexloge(ex)=52logex+1logex+logexlogeelogex=52logex+1logex+logex1logex=52logex=y(say)y+1y+y1y=52y(2y(1y))+2(1y)+2y2=5y(1y)2y22y3+22y+2y25y+5y2=02y39y2+7y2=0.

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