The natural logarithm function, $\ln$, is defined so that $e^{\ln x}=x$for any positive number$x$.
$\left(a\right)$Sketch a graph of the natural logarithm function.
$\left(b\right)$ Prove the identities
localid="1650331641178" $\ln ab=\ln a+\ln b$andlocalid="1650331643409" $\ln a^b=b\ln a$
(c) Prove thatlocalid="1650331645612" $\frac{d}{dx}\ln x=\frac{1}{x}$.
(d) Derive the useful approximation

localid="1650331649052" $\ln (1+x)\approx x$

which is valid when localid="1650331651790" $\left|x\right|\ll 1$. Use a calculator to check the accuracy of this approximation for localid="1650331654235" $x=0.1$and localid="1650331656447" $x=0.01.$

The logarithm base $e$is

$e^{\ln \left(x\right)}=x$

A plot of natural logarithm function graph as

The product logarithm is

$e^{\ln \left(x\right)}=x$

Where, $x=ab$as

$e^{\ln \left(ab\right)}=ab$

If $e^{\ln \left(a\right)}=a;e^{\ln \left(b\right)}=b$as

$\begin{array}{l}{e}^{\ln \left(ab\right)}=ab=e^{\ln \left(a\right)}{e}^{\ln \left(b\right)}\\ \to e^{\ln \left(ab\right)}=e^{\ln \left(a\right)+\ln \left(b\right)}\\ \ln \left(ab\right)=\ln \left(a\right)+\ln \left(b\right)\end{array}$

For exponents as,

$e^{\ln \left(x\right)}=x$

If $x=a^b$is

$e^{\ln \left(a^b\right)}=a^b$

where,$e^{\ln \left(a\right)}=a$is

$\begin{array}{l}{e}^{\ln \left(a^b\right)}={\left(e^{\ln \left(a\right)}\right)}^b=e^{b\ln \left(a\right)}\\ \ln \left(a^b\right)=b\ln \left(a\right)\end{array}$

Using implicit differentiation as

$e^{\ln \left(x\right)}=x$

Taking derivative on both sides as

role="math" localid="1650332642512" $$\begin{gathered} \frac{d}{dx}{e}^{\ln \left(x\right)}=\frac{d}{dx}x \\ \begin{array}{c}\frac{d}{dx}{e}^{\ln \left(x\right)}=1\end{array} \end{gathered}$$

Using exponential as

$\begin{array}{l}\frac{d}{dx}{e}^y=e^y\frac{d}{dx}y\\ e^{\ln \left(x\right)}\frac{d}{dx}\ln \left(x\right)=1\end{array}$

Where$e^{\ln \left(x\right)}=x$as

$\begin{array}{r}x\frac{d}{dx}\ln \left(x\right)=1\\ \frac{d}{dx}\ln \left(x\right)=\frac{1}{x}\end{array}$

By Taylor's formula

$f\left(x\right)=f\left(x_0\right)+{\left.\frac{df}{dx}\right|}_{x_0}\left(x-x_0\right)+\dots$

If $\ln (1+x)\text{at}{x}_0=0$as

$\begin{array}{l}\ln (1+x)=\ln (1+0)+{\left.(x-0)\frac{d(\ln (1+x\left)\right)}{dx}\right|}_0\\ \ln (1+x)=0+{\left.\left(x\right)\frac{1}{1+x}\right|}_0=\left(x\right)\frac{1}{1+0}\\ \ln (1+x)=x\end{array}$

Where $x=0.1$

$\ln (1+0.1)=\ln \left(1.1\right)=0.09531$

Which is nearly close to $0.1.$

Where $x=0.01$

$\ln (1+0.01)=\ln \left(1.01\right)=0.00995$

The approximation is nice here.