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Long Numbers and Logarithms

```Date: 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/
```
Associated Topics:
Middle School Logarithms

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