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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
$${Q}.\mathrm{210956} \\ $$$${im}\:{read}\:{leithold}\:{book}\:{again}\:,\:{in}\:{this}\:{book}\:: \\ $$$$\left.\mathrm{1}\right\}{define}\::\:{ln}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {dx}/{x}\:\:\:\:\:\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right\}{define}\::\:{ln}\left({e}\right)=\mathrm{1}=\int_{\mathrm{1}} ^{\:{e}} {dx}/{x} \\ $$$$\left.\mathrm{3}\right\}{define}\::\:{exp}\left({x}\right)={y}\:\Leftrightarrow\:{ln}\left({y}\right)={x} \\ $$$$\frac{{d}\left({ln}\left({u}\right)\right)}{{du}}=\frac{\mathrm{1}}{{u}}\:\Rightarrow\:\frac{{d}\left({ln}\left({u}\right)\right)}{{dx}}=\frac{{du}/{dx}}{{u}}\:\Rightarrow \\ $$$${u}={x}^{{r}} \:\Rightarrow\:\frac{{d}\left({ln}\left({x}^{{r}} \right)\right)}{{dx}}=\frac{{rx}^{{r}−\mathrm{1}} }{{x}^{{r}} }={r}×\frac{\mathrm{1}}{{x}}={r}×\frac{{d}\left({ln}\left({x}\right)\right)}{{dx}} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={rln}\left({x}\right)+{K}\:\Rightarrow\:{x}=\mathrm{1}\:\Rightarrow\:{K}=\mathrm{0} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={r}×{ln}\left({x}\right)\:\:\:\:\:\:\forall{x}>\mathrm{0}\:,\:\forall{r} \\ $$$${get}\:\:{x}={e}\:\:\Rightarrow\:{ln}\left({e}^{{r}} \right)={r}×{ln}\left({e}\right)={r}\:\Rightarrow\:{exp}\left({r}\right)={e}^{{r}} \\ $$$$\Rightarrow\:{exp}\left({x}\right)={e}^{{x}} ={y}\:\:\:\:\:\forall{x} \\ $$$${log}_{{e}} \left({e}^{{x}} \right)={log}_{{e}} \left({y}\right)={x}\:\:\:{and}\:\:\:{define}:\:{ln}\left({y}\right)={x} \\ $$$$\Rightarrow\Rightarrow\Rightarrow{ln}\left({y}\right)={log}_{{e}} \left({y}\right)={x} \\ $$

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