Q. 2.13 Fun with logarithms.a Simplify t... [FREE SOLUTION]

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An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 2: Q. 2.13 (page 62)

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

Short Answer

Expert verified

Part $(a)$

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

Part $(b)$

$(b)$ The equation $\ln 鈦� (a + b) 鈮� \ln 鈦� (a) + \frac{b}{a}$ is proved.

Step by step solution

Step: 1 Simplify steps: (part a)

We know that,

$\begin{matrix} e^{a \ln 鈦� (b)} \end{matrix}$

The logarithm rule is

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

Step: 2 Equating part: (part b)

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

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

Using Taylor formula as

$\begin{array}{l} f (x) = f (x_{0}) + {\frac{d f}{d x}|}_{x_{0}} (x 鈭� x_{0}) + 鈥� \\ \ln 鈦� (1 + \frac{b}{a}) = \ln 鈦� (1 + x) \end{array}$

Step: 3 Proving part: (part b)

We have,

$\ln 鈦� (1 + x) at x_{0} = 0$ as,

$\begin{array}{l} \ln 鈦� (1 + x) 鈮� \ln 鈦� (1 + 0) + {(x 鈭� 0) \frac{d (\ln 鈦� (1 + x))}{d x}|}_{0} \\ \ln 鈦� (1 + x) 鈮� 0 + {(x) \frac{1}{1 + x}|}_{0} = (x) \frac{1}{1 + 0} \\ \ln 鈦� (1 + x) 鈮� x \\ \ln 鈦� (1 + \frac{b}{a}) 鈮� \frac{b}{a} \\ \ln 鈦� (a + b) 鈮� \ln 鈦� (a) + \frac{b}{a} \end{array}$

Hence,the equation is proved.

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Short Answer

Step by step solution

Step: 1 Simplify steps: (part a)

Step: 2 Equating part: (part b)

Step: 3 Proving part: (part b)

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