• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q. 2.13

Expert-verified Found in: Page 62 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Fun with logarithms.$\left(a\right)$ Simplify the expression ${e}^{a\mathrm{ln}b}$. (That is, write it in a way that doesn't involve logarithms.)$\left(b\right)$ Assuming that $b<, prove that $\mathrm{ln}\left(a+b\right)\approx \left(\mathrm{ln}a\right)+\left(b/a\right)$. (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.)

Part $\left(a\right)$

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

Part $\left(b\right)$

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

See the step by step solution

## Step: 1 Simplify steps: (part a)

We know that,

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

The logarithm rule is

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

## Step: 2  Equating part: (part b)

Consider $b<and factor $a$ as

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

Using Taylor formula as

$\begin{array}{l}f\left(x\right)=f\left({x}_{0}\right)+{\frac{df}{dx}|}_{{x}_{0}}\left(x-{x}_{0}\right)+\dots \\ \mathrm{ln}\left(1+\frac{b}{a}\right)=\mathrm{ln}\left(1+x\right)\end{array}$

## Step: 3 Proving part: (part b)

We have,

$\mathrm{ln}\left(1+x\right)\text{at}{x}_{0}=0$ as,

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

Hence,the equation is proved. ### Want to see more solutions like these? 