Question

Fun with logarithms.
$\left(a\right)$ Simplify the expression$e^{a\ln b}$. (That is, write it in a way that doesn't involve logarithms.)
$\left(b\right)$ Assuming that $b<<a$, prove that $\ln (a+b)\approx (\ln a)+(b/a)$. (Hint: Factor out the $a$from the argument of the logarithm, so that you can apply the approximation of part $\left(d\right)$ of the previous problem.)

Short answer

Part $\left(a\right)$

$\left(a\right)$The expressions is $e^{a\ln \left(b\right)}=b^a.$

Part $\left(b\right)$

$\left(b\right)$The equation$\ln (a+b)\approx \ln \left(a\right)+\frac{b}{a}$is proved.

Step-by-step solution

Step: 1 Simplify steps: (part a)

We know that,

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

The logarithm rule is

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

Step: 2  Equating part: (part b)

Consider $b<<a$and factor $a$as

$\begin{array}{l}\ln (a+b)\\ \ln (a+b)=\ln \left[a\left(1+\frac{b}{a}\right)\right]\\ \ln (a+b)=\ln \left(a\right)+\ln \left(1+\frac{b}{a}\right)\end{array}$

Using Taylor formula as

$\begin{array}{l}f\left(x\right)=f\left(x_0\right)+{\left.\frac{df}{dx}\right|}_{x_0}\left(x-x_0\right)+\dots \\ \ln \left(1+\frac{b}{a}\right)=\ln (1+x)\end{array}$

Step: 3 Proving part: (part b)

We have,

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

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

Hence,the equation is proved.