Is it possible to calculate logarithm without calculator and without logarithm table?

**dariyoosh** · August 26th 2010, 08:19 AM

Dear all,

I would like to ask a question about calculating logarithm in base 10. Currently I'm reading an interesting book entitled: "ESSENTIALS OF PLANE TRIGONOMETRY AND ANALYTIC GEOMETRY".

The first chapter of the book is dedicated to logarithm function, its fundemantal forumlas, etc. However, I always used calculator in order to find the logarithm of a given number (base 10). In the book, there was some method (not really clear for me) presenting how to find the logarithm value of a given number based on the logarithm table.

I never managed to master the logarithm table (very complicated and ambiguous for me). Besides, after a bit googling I saw that there are different type of logarithm table (for base 10) with different number of columns and the method is not always the same.

Is there any simple method, allowing to calculate the logarithm of a number (base 10) without using a calculator and without using the logarithm table (with desired number of decimals)?

Thanks in advance,

Dariyoosh

**HallsofIvy** · August 26th 2010, 10:22 AM

There are a number of "simple" ways to calculate lograrithms- but they are all extremely tedious.

One method is to use the inverse function to the logarithm. The inverse function to $log_a(x)$ , the logarithm base a, is the exponential function $a^x$ .
In particular, the inverse to the "common logarithm", base 10, is $10^x$ while the inverse to the "natural logarithm", base e, about 2.718..., is $e^x$ .

Calculating $y= log_a(x)$ is the same as solving the equation [itex]x= a^x[/itex]. And there are a number of different, but tedious and repetative, ways to do that.

The simplest is the "midpoint method". First, by trial and error, find two numbers, $x_0$ and $x_1$ such that $log_a(x_0)< y$ and $log_a(x_1)> y$ . Since the logarithm is a "continuous function", there must some x between $x_0$ and $x_1$ such that $log_a(x)= y$ . We don't know exactly where between that number is so just try the midpoint: try $x_2= (x_0+ x_1)/2$ . If $log_a(x_2)= y$ we are done! If not, either $log_a(x_2)> y$ or $log_a(x)< y$ . If the first, then we know that there must be a solution between $x_0$ and $x_2$ , if the second between $x_2$ and $x_1$ . So we just do the same thing again, choosing new value for $x_3$ between our current $x_2$ and either $x_0$ or $x_1$ . At each step, we have reduced the interval, in which we know the value must be, by 1/2.

A slightly faster, slightly more complicated method, is the "secant method". Given two values, $x_0$ and $x_1$ , such that $e^{x_0}< y$ and $e^{x_1}> y$ , instead of the midpoint, construct the equation of the line through the two points $(x_0, e^{x_0})$ and $(x_1, e^{x_1})$ . There is a standard formula for the line between two points: $y= \frac{y_1- y_0}{x_1- x_0}(x- x_0)+ y_0$ is the line between $(x_0, y_0)$ and $(x_1, y_1)$ . Solve that equation for $x_2$ , determine whether $log_a(x_2)$ is larger than or less than y, and repeat.

A still faster but more sophisticated method is "Newton's method" for solving equations.
The derivative of $a^x$ is $ln(a) a^x$ where "ln(a)" is the natural logarithm, base "e", of a. To solve the equation f(x)= 0 by Newton's method, choose any starting value, $x_0$ (so you don't have to find two starting values on either side of the true solution) and form $x_1=x_0+ \frac{f(x_0}{f'(x_0)}$ . Basically, the idea is that we construct the straight line tangent to y= f(x) at the point $(x_0, f(x_0))$ and solve that linear equation to get a better approximation. Using that new value of $x_1$ , repeat to get a still better approximation. Newton's method has the property that it tends to double the number of correct decimal places in the answer on every step.

For $f(x)= a^x- y$ , for fixed y, so that f(x)= 0 is equivalent to [itex]a^x- y= 0[/tex] or $a^x=y$ , $f'(x)= ln(a) a^x$ so the formula becomes $x_1= x_0+ \frac{a^{x_0}- y}{ln(a) a^{x_0}}$ . In particular, for the natural logarithm, a= e, ln(a)= ln(e)= 1 so that is just $x_1= x_0+ \frac{e^{x_0}- y}{e^{x_0}}$ . For the common logarithm, a= 10 and $ln(a)= ln(10)$ so the formula is $x_1= x_0+ \frac{10^{x_0}- y}{ln(10) 10^{x_0}}$ . Of course, once you have found $x_1$ you continue in exactly the same way to find $x_2$ , $x_3$ , etc. to whatever accuracy you want.

Another straight forward method of finding approximate values of difficult functions is to use the "Taylor's series" about x= 0 (also called the "MacLaurin series") for the function. The Taylor's series for ln(1+ x) (Since log(0) is not defined, we cannot directly find the series for ln(x) about x= 0) is $1+ x+ x^2/2+ x^3/3!+ x^4/4!+ \cdot\cdot\cdot$ . As an infinite series, that is exactly $ln(x)$ . Taking a finite number of terms gives an approximation with the more terms taken, the better the approximation.

Finally, I am informed that modern calculators and computers use the "CORDIC" algorithm, about which I admit I know nothing! You might want to google on "CORDIC".

**RADHAKRISHNAN.J.** · July 18th 2011, 06:34 AM

This method may help you.http://rkmath.yolasite.com/resources/lt.pdf

**chisigma** · July 18th 2011, 07:04 AM

The following procedure is valid for the computation of the 'natural logaritm' of a number, i.e. the logarithm in base e. The logarithm in any other base is the 'natural logaritm' multiplied by a constant. The computation is based on the series expansion...

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

... which converges 'quickly enough' for $-.25 < x < .5$ . Now if You have to compute $\ln r$ with $r>1.5$ the procedure is...

a) devide r by 2 k times until obtain $\rho= \frac{r}{2^{k}}\ , \ .75< \rho < 1.5$ ...

b) set $x= \rho-1$ and compute with (1) $\ln (1+x)$ ...

c) compute $\ln r = \ln (1+x) + k\ \ln 2$ ...

Kind regards

$\chi$ $\sigma$

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