<s>[INST] Problem regarding solving logarithm without a calculator I have a question regarding the answer provided from this question, which is solving $\log$ without a calculator.
This was one of the solutions and I am having a hard time understanding it.

$$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$ Let us apply it for computing the logarithm of $\frac 53 =1.666667$; this would correspond to $x=\frac 14$. So $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \frac{1}{5120}+\cdots\right)=\frac{3923}{7680}\approx 0.5108$$ Now, $$\frac {166}{100}=\frac 53\times\frac{249}{250}$$ So $$\log(1.66)=\log \left(\frac53\right)+\log \left(\frac{249}{250}\right)=\log \left(\frac53\right)+\log \left(1-\frac{1}{250}\right)\approx\log \left(\frac53\right)-\frac{1}{250}$$ $$\log(1.66)\approx 0.5108-0.0040=0.5068$$ while the exact value would be ... the same.
If you need less accuracy, just truncate the first part, that is to say
$$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \cdots\right)=\frac{49}{96}$$ $$\log(1.66)=\frac{49}{96}-\frac{1}{250}=\frac{6077}{12000}\approx 0.5064$$

And my first question is how does the equation $\frac{166}{100} = \frac{5}{3} \times \frac{249}{250} $ came out.
Second question is how does $\log(1- \frac{1}{250})$ becomes $-\frac{1}{250}$.
Thank you for your help.
 [/INST] There is  formula the solver knew very well and was comfortable with.

$\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......$

From this they conclude two things

$\ln \frac{1+x}{1-x}=\ln(1+x) - \ln (1-x) = (x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......) - (-x -\frac {x^2}2 - \frac {x^3}3- \frac {x^4}4 -......)=$
$2(x +\frac {x^3}3 + \frac {x^5}5 + ...... )$

And

If $|x|$ is very very small then all the $\frac {x^k}k$ terms are negligibly small and $\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + .... \approx x$.

They figure that as $ 1.66$ is close to $\frac 53$ and we can use $\frac 53 = \frac {1+\frac 14}{1-\frac 14}$ and use the second formula to approximate $\ln \frac 53$.  (Although why they didn't they use $\frac 53 = 1 + \frac 23$ and use the first formula, isn't clear to me.)
But as they didn't want $\ln \frac 53$ but $\ln 1.66$.  However they figured that as $ 1.66$ is smaller but very close to $\frac 53$ there must be some very small $e$ so that $1.66 = \frac 53(1-e)$.
If so, to solve $\ln 1.66=\ln\frac 53 (1-e)$ we get
$\ln \frac 53 + \ln (1-e)=$
$\ln \frac {1+\frac 14}{1-\frac 14} + \ln (1-e)$.
By the second formula we can figure out what $\ln \frac {1+\frac 14}{1-\frac 14}$ is.
And by the third approximation we can assume $\ln (1-e) \approx e$.
So what is $e$?
Well $1.66 = \frac 53(1-e)\implies e =\frac 1{250}$.
But they didn't actually do it that way.  They did:
$1.66 =(1.66\cdot \frac 35)\times \frac 53 =$
$\frac {166\cdot 3}{500}\times \frac 53 =$
$\frac {83\cdot 3}{250}\times \frac 53=$
$\frac {249}{250}\times \frac 53=$
$\frac 35(1-\frac 1{250})$.
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