How does the poor man’s log₁₀ calculator trick work?

Reddit Post 

https://www.reddit.com/r/explainlikeimfive/comments/fyh9qr/eli5_how_does_the_poor_mans_log₁₀_calculator/

So… This is an interesting one. During my high-school days, most of us here in India couldn’t afford a decent calculator. We all had one of those cheap ones that can do basic BODMAS and additionally, square roots. Log books were also rare and used to go out of stock pretty fast.

There is this brilliant trick that used to work perfectly to find the log₁₀ of any number upto 3 digits on this regular calculator:

1. Type the number.
2. Recursively take its square root 19 times.
3. Subtract one.
4. Multiply the result by 227697.

And miraculously, it gave the log₁₀(x) accurate upto 4/5 decimal places. Sometimes even more. How does this work?

My response:

Suppose x>0.

The limit of (x^a -1)/a a->0 from the positive side can be computed using l’Hospital’s rule.

lim (x^a -1)/a = lim ( exp( a ln(x)) -1)/a = lim ln(x) exp( a ln(x) )/1 = ln(x).

So, for small positive a,

(x^a -1)/a ≈ ln(x).

For log base 10, (x^a -1)/a/ln(10) ≈ log_10(x). 

If we choose a = 0.5^19 = 1.90735*10^-6, then 

log_10(x) ≈ 227695*(x^(0.5^19) -1).

If we choose a = 1/10^6 , then 

log_10(x) ≈ 434294*(x^(1/10^6) -1).