I’m clearly missing something here. How can the two to an exponent be bigger than the answer when you count all the digits in the answer versus the number of digits in the exponent?
10^1 is 2 digits long. 10^2 is 3 digits long. 10^9 (the largest 1-digit exponent) is 10 digits long. 10^99 (largest 2-digit exponent) is 100 digits long.
We can see that the length of the result grows faster than the length of the exponent. For 10^x, there will be x digits in the result, but the digit-length of x itself will be just CEIL(LOG10(x)).
The same thing applies for any positive base of your exponents still expressed in base 10 numerals. Using 10 also as our exponent base just makes it real easy to see in our common numbering system.
For exponents of 2 (i.e. 2^x expressed in base 10 numerals), the rate of growth is slower as the exponents increase. But by the time the exponent is hundreds of millions, the resulting number of digits is also mongo.
Are you mixing up decimal digits and base 2 digits? In base 2, that number is 136,279,841 digits long (and in fact, all of those digits are 1s). But numbers can be written shorter in base 10, less than a third the length, because 10 is a bit bigger than 2^3.
I’m not quite sure I understand the problem, but I’ll try to answer. 10^(.301) is very close to 2, so 10^(.301 times x) will be close to 2^(x). So, as you say, 2^7 is a bit more than 10^2 (7*.301= 2) (a better example is 2^10 which is very close to 10^3 (1024 versus 1000).
2^136,279,841 should be close to 10^(136,279,841*.301) - and it is
I am unsure of the moves between Base-2 and Base-10 here.
The book I have says 2^136,279,841 -1 results in an answer that is 41,024,320 digits long.
136,279,841 > 41,024,320
2^n seems to ALWAYS result in an answer that has at least the same number of digits in the answer or more than the exponent.
This seems to go the other way. 2^136,279,841 -1 results in an answer with far fewer digits than it started with in the exponent. How does that happen?
Yes, the number is longer in base 2 than in base 10.
2^1, 2^2, and 2^3 all have 1 digit. 2^4, 2^5, and 2^6 all have two digits, etc. The number of (base ten) digits is always less than the (base two) exponent.
It is a very useful rule of thumb that: 2^{10} = 1024 \simeq 1000 = 10^3
0.3 times the exponent of the power of 2 and you are very close to the equivalent power of 10. (Or just dividing by 3 isn’t terrible most of the time.) The converse is useful as well. Every 3 decades of range needs 10 additional bits to represent.
I used to drill this into my various computer science classes. (I would invariably get asked if they could bring a calculator into an exam to calculate trivial binary arithmetic results. All you have to remember is that 2^{10} = 1024.)
In general, the conversion is of the form: {ln(2) \over ln(10) }= .301
Simple rule of thumb I used when working with binary back in the day. 2^10 is 1024, or roughly 1,000 or 10^3; Hence called 1K.
So for 2^X, divide X by 10, multiply by 3, that’s the approximate number of digits. Of course that extra 24 (1,000 vs 1,024) will compound, but for back of the envelope order of accuracy, pretty close. (I.e. 2^30 is about 10^9)
136,279,84.1 * 3 = 40,883,952 digits. Pretty darn close.