Is it possible to calculate log10 x without using log?

Question

Is it possible to calculate $\log_{10} x$ without using $\log_{10}$? I'm interested because I'm working with a framework that has some simple functions, but log is not one of them.

The specific platform is capable of doing addition, subtraction, multiplication and division. I can write a formula of a finite, predefined length; e.g. while loops or for loops that continue indefinitely are not available. It's a black-box project with an interface capable of creating formulas, so basically, I need to write the expression to calculate $\log_{10} x$ in one line, and I can't write my own method or something to do more dynamic calculations.

An approximation of $\log_{10} x$ is perfectly acceptable.

Answer 1

You can use a Taylor polynomial to first roughly compute $\ln n$. $$\ln n\approx0+\dfrac{1}{1!}(n-1)+\dfrac{(\frac{-1}{n^2})}{2!}(n-1)^2+\dfrac{(\frac{2}{n^3})}{3!}(n-1)^3+\dfrac{(\frac{-6}{n^4})}{4!}(n-1)^4+\dfrac{(\frac{24}{n^5})}{5!}(n-1)^5$$ Then use the fact that $\ln 10 \approx2.302585092994046$, so divide $\ln n$ by that to get $\log_{10} n$

Answer 2

Many platforms will at least have a function for the natural logarithm ($\ln$).

Note that $\log_{10}(x) = \ln(x)/\ln(10)$.

Answer 3

Write $x=a\times 10^b$ with $0<a<1$. So, $$\log_{10}x=b+\log_{10}(a)=b+\frac{\log (a)}{\log (10)}$$ Now, use the very fast convergent series expansion $$\log\Big(\frac{1+x}{1-x}\Big)=2\Big(\frac{x}{1}+\frac{x^3}{3}++\frac{x^5}{5}+\cdots\Big)$$ using $a=\frac{1+x}{1-x}$ that is to say $x=\frac{a-1}{a+1}$ and you know that $\log(10)\approx 2.30259$.

Answer 4

The CORDIC algorithms may help here (used by most calculators working in BCD in the old days because requiring only additions and shifts).

It uses an array $\rm{L}$ of precalculated logarithms in base $10$ ($\log$ means $\log_{10}$ here) and
$x$ will be supposed written using digits (with a decimal point) :

L= [log(2), log(1.1), log(1.01), ... , log(1.000001)];

The relation $\;\displaystyle \log(x) = \log\left(x\,10^{−n}\right) + n\;\;$ is used to get all positive real values $x$ in the range $(1,10]\;$ ($n$ is the number of shifts of the decimal points required for this purpose, $10^{−n}$ really means "right-shift of $n$" for $n>0$).

The algorithm may then be applied to $\;x\mapsto x\,10^{−n}$ :

k= 0; y= 0; p= 1;  //p is the partial product
while (k <= 6)
{
   while (x >= p+p*10^(-k))
   {
      y= y+L[k];       // L[k] = log(1+10^{-k})
      p= p+p*10^(-k);
   }
   k= k+1;
}
return y;

The idea is that for $$x=(1+1)^{n_0}\,(1+10^{-1})^{n_1}\,(1+10^{-2})^{n_2}\cdots (1+10^{-6})^{n_6}(1+\epsilon)$$ we will get $$\log(x)=n_0\log(1+1)+n_1\log(1+10^{-1})+\cdots+n_6\log(1+10^{-6})$$ and the precision should be near $7$ digits with the upperbound $6$ (more terms means more precision).

Answer 5

Let x = 1545 and proceed as follows:

1. Divide x by base 10 until x becomes smaller than base 10. 1545/10 = 154.5 154.5 / 10 = 15.45 15.45 / 10 = 1.545 As we divide 3 times the value of x by 10, the whole digit of our logarithm will be 3.

2. Raise x to the tenth power: 1.545 ^ 10 = 77.4969905 ...

3. Repeat the procedures from step 1.

As we divide only 1 time, the next digit of our logarithm will be 1.

4. Repeat step 2.

7.74969905 ... ^ 10 = 781.354.964,875286221 ...

5. Repeat step 1.

Since we divide x by 10 by 8 times, the next digit in our log will be 8.

And so on, as many digits as you want.