4
$\begingroup$

I have a calculator that does not have antilog function. All it has is log to base 10 and natural log functions.

I was wondering if it is possible to calculate antilog using the log to base 10 function. Can this be done ? I am only concerned about log to base 10 and antilog to base 10.

$\endgroup$
6
  • 6
    $\begingroup$ I'm not sure what an antilog is --- isn't it just your $e^x$ button or $10^x$? if $\log_b(x) = y$, then $b^y = x$. $\endgroup$
    – Newb
    Nov 28, 2013 at 17:37
  • $\begingroup$ What brand & model? It will be easier to help, since most major vendors put manuals online. $\endgroup$ Nov 28, 2013 at 17:38
  • 1
    $\begingroup$ I have Casio fx-82 mx $\endgroup$
    – An SO User
    Nov 28, 2013 at 17:39
  • 5
    $\begingroup$ stop calling exponential anti logarithm and look into manual or on the labels of the buttons... $\endgroup$
    – V-X
    Nov 28, 2013 at 17:40
  • 1
    $\begingroup$ Ok, inverse of ln is shift-ln (or $e^x$), and inverse of decimal log is shift-log (or $10^x$). If you want power or exponentiation or whatever you call it, it's shift-$\times$. $\endgroup$ Nov 28, 2013 at 17:41

3 Answers 3

6
$\begingroup$

Antilog is just another name for exponentiation. I'm sure your calculator has exponentiation! Or if it doesn't have, you can do repeated multiplication. In your case Antilog to base 10 is 10 raised to power something. And btw, I also have Casio fx-82 ms, and it has exponentiation.

$\endgroup$
2
  • 7
    $\begingroup$ Repeated multiplication is not much helpful for finding $10^{2.188377266432}$, though. $\endgroup$ Nov 28, 2013 at 17:41
  • 1
    $\begingroup$ Yes, very true! I was just trying to exclaim. $\endgroup$
    – Shubham
    Nov 28, 2013 at 17:43
5
$\begingroup$

there is a $y^x$ or $pow$ button. (Obviously, it is not on the dumb +- caluclators)

$\endgroup$
1
  • $\begingroup$ If the calculator is too dumb to have exponentiation, it will be too dumb to have logs. $\endgroup$ Apr 3, 2019 at 3:37
1
$\begingroup$

Assuming, as you say, that your calculator does not have the 10^x function, the approach depends on how much accuracy you need. I advise to memorize the logarithm of e, 0.4342944819. For some methods you will need all these digits, for other methods, you only need the first two digits, but they repeat, so just retain in your head 0.4343.

Look at the number whose antilog you need. The integer part is just the power of ten, so write that down immediately. The fractional part is where you will have the fun. The following method is from the pre-calculus method for calculating logarithms, based on the bisection algorithm: create a table with four columns: x(lower limit), x(upper limit), antilog(lower limit), antilog(upper limit). The first row will contain 0, 1, 1, 10. Calculate the mean of the first two columns and the square root of the product of the third and forth columns (also known as the geometric mean). If your x value is greater than the mean, replace the lower limit for x with the mean and the lower limit for the antilog with the geometric mean. Otherwise, replace the upper limits. Now repeat this process. You will find the first two columns sandwiching down on your x value and the third and fourth column sandwiching down on your antilog. I leave it as an exercise for the student to use this exact same table for computing logarithms. Wow! Back in the 17th century they must have really enjoyed calculating square roots because you must calculate the square root to at least the same number of digits as your required answer.

A second method is based on the slope of the antilog curve. The slope is just the antilog divided by the log of e. For this work, 0.43 is close enough. So suppose you want the antilog of 0.2. You need an initial guess, and the closer the guess is to your correct answer, the less work you will do. From my experience with slide rules over the last 50 years, I remembered the approximate values of the antilog for 0, 0.1, 0.2, ..., 0.9, 1.0. You probably already know the first, the middle, and the last of these. Taking 1.6 as the initial guess for the antilog, I use the log button to get log(1.6)=0.20411. The guess was a bit in the high side. So subtract 0.00411*1.6/0.43=0.01529. This gives a revised guess of 1.58470. If you take the log of this, you get 0.19994897, which is pretty close to what you need. However, for greater accuracy, you can repeat this process with the revised guess. To save time, you might calculate the factor 1.6/0.43 and store it in the calculator memory, if you have one.

A third approach is based on the calculus of limits. For this, you need to know that 0.43 constant to full precision, but the method is quick and dirty. Suppose you need the antilog of x. Divide x by 1024. Divide that by 0.4343. Add 1. Square this ten times. Example: antilog(0.9). I got 7.926 instead of the correct 7.943, partly because I didn't remember the full 12 digits of 0.43, but it is so quick and easy, you might as well know how to do it. For x closer to 0 the algorithm is more accurate, but the calculator does not carry enough digits to accurately do the calculation. Try dividing by 128 and doing the square root only seven times. It is left as an exercise for the student to learn to use this technique to calculate logarithms.

Other approaches are by power series and continued fractions, but I have no desire to inflict additional torture on you.

My recommendation is that you invest in a new calculator! If you need a graphic calculator, the HP Prime is excellent. Otherwise the WP-34. Instead of getting the hardware with shoddy keys that are prone to failure, get the app for your cell phone.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.