What's The Formula For Calculating Large Exponents? - Physics Forums

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What's the formula for calculating large exponents?
  • Context: High School
  • Thread starter Thread starter slakedlime
  • Start date Start date Nov 1, 2007
  • Tags Tags Exponents Formula
Click For Summary

Discussion Overview

The discussion revolves around the calculation of large exponents, specifically evaluating \(16^{198}\) without the use of a calculator. Participants explore various methods for approximating the value and finding specific digits, including the last two and six digits, while considering mathematical techniques such as logarithms and modular arithmetic.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant questions the feasibility of evaluating \(16^{198}\) without a calculator and suggests it may be more reasonable to find an approximation or the last few digits instead.
  • Another participant proposes using logarithmic approximations, noting that \( \log_{10}(2) \) is approximately 0.3, which allows for quick calculations of powers of 2 and 10.
  • A different participant expresses interest in finding the last six digits and suggests that modular mathematics might be relevant for this purpose.
  • Some participants argue about the practicality of evaluating the expression without a calculator, with one stating that modern computers can compute it quickly, while another emphasizes the challenge of doing so by hand.
  • One participant suggests using binary representation as an alternative approach.
  • A later reply provides a detailed modular arithmetic calculation to find the last two digits of \(16^{198}\), showing the steps involved in the computation.
  • Some participants express confusion about specific steps in the modular arithmetic process, indicating a need for clarification on the calculations presented.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for calculating \(16^{198}\) without a calculator. There are multiple competing views on the feasibility of direct evaluation versus approximation and modular arithmetic techniques.

Contextual Notes

Participants highlight the complexity of the calculations involved and the limitations of performing them without computational tools. There is also mention of the need for clarity in the modular arithmetic steps, indicating that some assumptions may not be fully articulated.

Who May Find This Useful

This discussion may be useful for students preparing for assessments involving large exponent calculations, as well as those interested in mathematical techniques for approximating values and using modular arithmetic.

slakedlime Messages 74 Reaction score 2 [SOLVED] What's the formula for calculating large exponents? Hi! I came across the following sum and was wondering if there was some kind of formula for solving it: 16^198 Is it some kind of binomial expansion? Do you calculate the last digits first, advancing to the previous ones later? I can't use a calculator. Please, please help. Sums like this might come in an assessment exam I'm sitting. Thank you very much! Last edited: Nov 1, 2007 Mathematics news on Phys.org
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Gib Z Homework Helper Messages 3,341 Reaction score 7 It wants you to evaluate 16^(198) without a calculator? Its approximately 2.60469314 \cdot 10^{238}, even with an efficient method it would take an enormous amount of time to evaluate, you sure you don't mean approximate, or find the last 5 digits of it? uart Science Advisor Messages 2,797 Reaction score 21 Yeah it's not actually solving it but I can make a very rough approximation because I happen to remember that log_10(2) is pretty close to 0.3. Keeping this approx figure in your head let's you do quick and nasty power of 2 to power of 10 (or visa versa) conversions without a calculator. So in this case 16^{198} = 2^{4*198} \simeq 10^{0.3 * 4 * 198}, which gets you an order of magnetude calcuation of 10^238. Last edited: Nov 2, 2007 slakedlime Messages 74 Reaction score 2 I can do the approximations, but is there any way to find the last 6 digits? The question asked for both the approximation and the last 6 digits, so I thought perhaps the method would be similar and that there would be a formula. Does this have something to do with modular mathematics? Thank you for your help. :) Zurtex Science Advisor Homework Helper Messages 1,118 Reaction score 1
Gib Z said: It wants you to evaluate 16^(198) without a calculator? Its approximately 2.60469314 \cdot 10^{238}, even with an efficient method it would take an enormous amount of time to evaluate, you sure you don't mean approximate, or find the last 5 digits of it?
Why would it take an enormous amount of time to evaluate? Just multiply it out... 260469313784369307581244210575049132700967121965465162515478820772032704602251252793805945346545089482145699632555985954917531314614037698451693595794... (Edit I've removed the last load of digits not to make it too easy for you). Takes fractions of seconds on modern computers. And yeah, doing it via modular arithmetic is even easier. Because if you want to know the last 6 digits you can just take it mod 1'000'000, so you can just keep knocking off any digits larger than the first 6. Last edited: Nov 2, 2007 Moo Of Doom Messages 365 Reaction score 1
Zurtex said: Takes fractions of seconds on modern computers.
You didn't read the "without a calculator" part, didn't you? Zurtex Science Advisor Homework Helper Messages 1,118 Reaction score 1
Moo Of Doom said: You didn't read the "without a calculator" part, didn't you?
Well I didn't use a calculator :P But yeah, it looks like you have to do some clever modular arithmetic, it'd pretty easy mod 10, mod 100 or perhaps even mod 1'000 by hand, but it's a bit of a pain mod 1'000'000. I'll think on a bit and see if I come up something (certainly nothing that immediately came to mind worked). Count Iblis Messages 1,858 Reaction score 8 I would suggest to give the answer in the binary number system :smile: slakedlime Messages 74 Reaction score 2 Thank you all for trying. Forget 6 digits. My exam's tomorrow afternoon, so if it'd be easier to find the last two digits without using a calculator, could you guys show me how? I'd just like to have an idea even if I can't answer. Count Iblis Messages 1,858 Reaction score 8 16^198 = 16^(2*3^2 * 11) All computations mod 100 below: 16^2 = 256 = 56 16^(2*3) = 56^3 = 16 16^(2*3^2) = 16^3 mod(100) = 96 = -4 16^(2*3^2*11) = (-4)^11 =-2^22 now 2^10 = 1024 = 24 so 2^20 = 24^2 = 576 = 76 2^22 = 2^20 * 4 = 76*4 = 4 ------> 16^198 = -4 = 96 slakedlime Messages 74 Reaction score 2
Count Iblis said: 2^22 = 2^20 * 4 = 76*4 = 4 ------> 16^198 = -4 = 96
I didn't quite catch that part. Zurtex Science Advisor Homework Helper Messages 1,118 Reaction score 1
slakedlime said: I didn't quite catch that part.
Reducing it down to the results rather than the calculations: 16^198 = 16^(2*3^2 * 11) 16^(2*3^2*11) = (-4)^11 =-2^22 2^22 = 4 16^(198) = -4 -4 is the same as 96, mod 100.

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