How to show the approximation $\frac{x_{k+1} - x_k}{x_k} \approx \log x_{k+1}-\log x_k$

Question

In the book of Bouchaud and Potters 2003, The theory of financial risk and derivatives, at page page 90, it is stated that

$$ \frac{x_{k+1} - x_k}{x_k} \approx \log x_{k+1}-\log x_k $$ if $x_{k+1} - x_k$ is much smaller than $x_k$. $\{x_k\}$ is a sequence.

However, I can’t seem to derive the approximation. I tried taking to log-than-exponential of both sides and got $$ \exp{\left(\log{\left(\frac{x_{k+1}}{x_k} - 1\right)}\right)} $$ and considered the limit where $\frac{x_{k+1}}{x_k} \to 1$, but couldn’t really get anywhere.

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Please note that when the fraction in LHS is converted to log, the problem becomes

$$ \log ({x - 1}) \quad as \quad x\to 1. $$

Answer 1

Note that $\log x_{k+1}-\log x_k=\log(x_{k+1}/ x_k)=\log(\frac{x_{k+1}-x_k}{x_k}+1)$ and that $\log(1+u)=u-u^2/2+u^3/3-\ldots\approx u$.

Answer 2

Start from the right-hand side and rewrite it as $$ \log x_{k+1} - \log x_k = \log \frac{x_{k+1}}{x_k} = \log \left(1+\frac{x_{k+1}-x_k}{x_k}\right) $$ then use the first order expansion of the logarithm $\log x+1 \approx x$ if $x\ll 1$. and obtain $$ \log x_{k+1} - \log x_k \approx \frac{x_{k+1}-x_k}{x_k} $$