Long Numbers and LogarithmsDate: 10/11/2002 at 20:05:43 From: Richard Subject: Numbers, exponents Hi, I have a problem figuring out how many terms are there when I raise a number to another large number or a large exponent. Is there a formula for these problems? Example: (12)^5 = 248832 There are 6 terms in the answer. (12)^6 = 2985984 There are 7 terms in the answer. (12)^7 = 35831808 There are 8 terms in the answer. (12)^(214) How many terms will there be? How about (123)^214? How many terms are there? Date: 10/12/2002 at 12:50:26 From: Doctor Ian Subject: Re: Numbers, exponents Hi Richard, Usually if you want to work with large numbers of digits, you have to use logarithms. For example, consider 12^5 = 12 * 12 * 12 * 12 * 12 log(12^5) = log(12 * 12 * 12 * 12 * 12) = log(12) + log(12) + log(12) + log(12) + log(12) = 5 * log(12) = 5 * 1.08 = 5.40 So you'd expect to see 6 digits. Now, suppose we look at a higher exponent, like 20. log(12^20) = 20 * log(12) = 20 * 1.08 = 21.6 So here, we'd expect to see 22 digits, not 21. And log(12^214) = 214 * log(12) = 214 * 1.08 = 231.12 which would have 232 digits. Because this is close to 231, it's a good idea to use more digits in the approximation of the logarithm, to make sure we're not over-estimating: log(12^214) = 214 * log(12) = 214 * 1.07918 = 230.94452 So we'd really expect to see 231 digits. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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