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Given "$x = \log 8$", it is very easy to rewrite the expression as "$10^x = 8$", which cannot easily be solved for by hand. However, if I plug "$x = \log 8$" into my calculator, I get "$x = 0.903089986992$".

So How Does It Know?
Is there some sort of logarithmic formula implemented by calculators, or does it really just brute-force the value for all of those decimal places?

Note:
I develop apps as a hobby, so if said formula involves loops or binary-operators, don't feel overly pressured to explain how they work; I already understand them.

  • It most likely uses a taylor series expansion for the logarithm to approximate it. – Edward Evans Apr 03 '16 at 17:29
  • you can have a look to this post, it might be helpful for your question http://math.stackexchange.com/questions/698384/approximating-logs-and-antilogs-by-hand/919287#919287 – user3209698 Apr 03 '16 at 17:32
  • As said by Ed_4434, Taylor Series are likely used, as well as probably the idea that $\log_b{a}=\frac{\ln a}{\ln b}$. – Ethan Hunt Apr 03 '16 at 18:04

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As pointed out in the comments, it probably uses the Taylor Series, which gives us $$ \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots $$ However, more insightful answers are given here.

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