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Q-210956-im-read-leithold-book-again-in-this-book-1-define-ln-x-1-x-dx-x-x-gt-0-2-define-ln-e-1-1-e-dx-x-3-define-exp-x-y-ln-y-x-d-ln-u-du-1-u-d-ln-u-dx




Question Number 210967 by mahdipoor last updated on 24/Aug/24
Q.210956  im read leithold book again , in this book :  1}define : ln(x)=∫_1 ^( x) dx/x      x>0  2}define : ln(e)=1=∫_1 ^( e) dx/x  3}define : exp(x)=y ⇔ ln(y)=x  ((d(ln(u)))/du)=(1/u) ⇒ ((d(ln(u)))/dx)=((du/dx)/u) ⇒  u=x^r  ⇒ ((d(ln(x^r )))/dx)=((rx^(r−1) )/x^r )=r×(1/x)=r×((d(ln(x)))/dx)  ⇒ ln(x^r )=rln(x)+K ⇒ x=1 ⇒ K=0  ⇒ ln(x^r )=r×ln(x)      ∀x>0 , ∀r  get  x=e  ⇒ ln(e^r )=r×ln(e)=r ⇒ exp(r)=e^r   ⇒ exp(x)=e^x =y     ∀x  log_e (e^x )=log_e (y)=x   and   define: ln(y)=x  ⇒⇒⇒ln(y)=log_e (y)=x
Q.210956imreadleitholdbookagain,inthisbook:1}define:ln(x)=1xdx/xx>02}define:ln(e)=1=1edx/x3}define:exp(x)=yln(y)=xd(ln(u))du=1ud(ln(u))dx=du/dxuu=xrd(ln(xr))dx=rxr1xr=r×1x=r×d(ln(x))dxln(xr)=rln(x)+Kx=1K=0ln(xr)=r×ln(x)x>0,rgetx=eln(er)=r×ln(e)=rexp(r)=erexp(x)=ex=yxloge(ex)=loge(y)=xanddefine:ln(y)=x⇒⇒⇒ln(y)=loge(y)=x

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