Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

Fun with logarithms.
$(a)$ Simplify the expression $e^{a \ln b}$ . (That is, write it in a way that doesn't involve logarithms.) $(b)$ 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 $(d)$ of the previous problem.)

Part $(a)$

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

Part $(b)$

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step: 1 Simplify steps: (part a)

We know that,

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

The logarithm rule is

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

Step: 2 Equating part: (part b)

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

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

Using Taylor formula as

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

Step: 3 Proving part: (part b)

We have,

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

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

Hence,the equation is proved.

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