Question Number 210967 by mahdipoor last updated on 24/Aug/24
$${Q}.\mathrm{210956} \\ $$$${im}\:{read}\:{leithold}\:{book}\:{again}\:,\:{in}\:{this}\:{book}\:: \\ $$$$\left.\mathrm{1}\right\}{define}\::\:{ln}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {dx}/{x}\:\:\:\:\:\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right\}{define}\::\:{ln}\left({e}\right)=\mathrm{1}=\int_{\mathrm{1}} ^{\:{e}} {dx}/{x} \\ $$$$\left.\mathrm{3}\right\}{define}\::\:{exp}\left({x}\right)={y}\:\Leftrightarrow\:{ln}\left({y}\right)={x} \\ $$$$\frac{{d}\left({ln}\left({u}\right)\right)}{{du}}=\frac{\mathrm{1}}{{u}}\:\Rightarrow\:\frac{{d}\left({ln}\left({u}\right)\right)}{{dx}}=\frac{{du}/{dx}}{{u}}\:\Rightarrow \\ $$$${u}={x}^{{r}} \:\Rightarrow\:\frac{{d}\left({ln}\left({x}^{{r}} \right)\right)}{{dx}}=\frac{{rx}^{{r}−\mathrm{1}} }{{x}^{{r}} }={r}×\frac{\mathrm{1}}{{x}}={r}×\frac{{d}\left({ln}\left({x}\right)\right)}{{dx}} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={rln}\left({x}\right)+{K}\:\Rightarrow\:{x}=\mathrm{1}\:\Rightarrow\:{K}=\mathrm{0} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={r}×{ln}\left({x}\right)\:\:\:\:\:\:\forall{x}>\mathrm{0}\:,\:\forall{r} \\ $$$${get}\:\:{x}={e}\:\:\Rightarrow\:{ln}\left({e}^{{r}} \right)={r}×{ln}\left({e}\right)={r}\:\Rightarrow\:{exp}\left({r}\right)={e}^{{r}} \\ $$$$\Rightarrow\:{exp}\left({x}\right)={e}^{{x}} ={y}\:\:\:\:\:\forall{x} \\ $$$${log}_{{e}} \left({e}^{{x}} \right)={log}_{{e}} \left({y}\right)={x}\:\:\:{and}\:\:\:{define}:\:{ln}\left({y}\right)={x} \\ $$$$\Rightarrow\Rightarrow\Rightarrow{ln}\left({y}\right)={log}_{{e}} \left({y}\right)={x} \\ $$