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


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Question Number 210967 by mahdipoor last updated on 24/Aug/24

$Q . 210956$ $i m r e a d l e i t h o l d b o o k a g a i n, i n t h i s b o o k :$ $1} d e f i n e : l n (x) = \int_{1}^{x} d x / x x > 0$ $2} d e f i n e : l n (e) = 1 = \int_{1}^{e} d x / x$ $3} d e f i n e : e x p (x) = y \Leftrightarrow l n (y) = x$ $\frac{d (l n (u))}{d u} = \frac{1}{u} \Rightarrow \frac{d (l n (u))}{d x} = \frac{d u / d x}{u} \Rightarrow$ $u = x^{r} \Rightarrow \frac{d (l n (x^{r}))}{d x} = \frac{{r x}^{r - 1}}{x^{r}} = r \times \frac{1}{x} = r \times \frac{d (l n (x))}{d x}$ $\Rightarrow l n (x^{r}) = r l n (x) + K \Rightarrow x = 1 \Rightarrow K = 0$ $\Rightarrow l n (x^{r}) = r \times l n (x) \forall x > 0, \forall r$ $g e t x = e \Rightarrow l n (e^{r}) = r \times l n (e) = r \Rightarrow e x p (r) = e^{r}$ $\Rightarrow e x p (x) = e^{x} = y \forall x$ ${l o g}_{e} (e^{x}) = {l o g}_{e} (y) = x a n d d e f i n e : l n (y) = x$ $\Rightarrow\Rightarrow\Rightarrow l n (y) = {l o g}_{e} (y) = x$
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