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Can you find the logarithm and antilogarithm of a number withou using logtable, calculator or computer?
Is there any formula or something to find out log and antilog of a number?
I want this to calculate:
X^Y
where X and Y are both fractions.
8 Answers
- bpiguyLv 71 decade agoFavorite Answer
Your answer is yes. There are series expansions for both Napierian (base e) and common (base 10) logs and antilogs. It appears that Phoenix (see above) has one.
Kepler used logs for his calculations 400 years ago, and they didn't have computers or calculators then -- but they did have tables, and you can interpolate using 5-place tables.
I'd have to look in a calculus book to find the series expansions, but for what you want to do, I don't think you need them.
You want to calculate X^Y, where both X and Y are fractions. Suppose Z = X^Y, where X = A/B. Then
Z = X^Y = (A/B)^Y = (A^Y)/(B^Y)
log Z = log[(A^Y)/(B^Y)] = log(A^Y) - log(B^Y)
log Z = Y log A - Y log B = Y(log A - log B)
Then, to get Z, all you do is take the antilog.
Of course, maybe this is <not> what you want to do. If you <really> want to do it using the series expansions, they're available. Here's one (I don't know if this is what Phoenix used):
(1/2) log[(1+z)/(1-z)] = z + (z^3)/3 + (z^5)/5 + ...
- Anonymous1 decade ago
Yes you can, but it is complicated:
1. Natural Logarithm of a number is given by the log-series:
ln (1 + x) = x - (x^2)/2 + (x^3/3) - (x^4/4) + .........
plug in the value for 'x' as required and evaluate RHS upto infinite terms, lol... j/k (you can evaluate upto 10 terms to be on the safe side)
2. Antilog of a number 'x' is exponential 'e' raised to the power of 'x', where e = 2.17828. To evaluate power of a real number, use the binomial theorem.
- 1 decade ago
yeah.. i think.. cant u just convert it over to the log of 10 with the constant being X and have the Y move to the front of the logarithm and then if u have a different base then just convert...
i dont know.. Calculus was like last year.. thats a whole summer inbetween..:)
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- Anonymous1 decade ago
Yes, but it is a secret...