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Q. 2.13

Expert-verifiedFound in: Page 62

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 $b<<a$, prove that $\mathrm{ln}(a+b)\approx (\mathrm{ln}a)+(b/a)$. (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.)

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

We know that,

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

The logarithm rule is

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

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

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

Using Taylor formula as

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

We have,

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

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

Hence,the equation is proved.

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