Finding Number of digits in an exponent without using a calculator?

[Firas]

What is a good way / formula (if there is) to find the number of digits in an exponent without using a calculator?

Is there a pattern? For example, how many digits are there in 2^130? I know that you can use logarithms but keep in mind, I cannot use a calculator.

 

[Deleted member 4993]

You can provide upper bound and lower bound by denoting:

2³ < 10 < 2⁴

 

[Firas]

You can provide upper bound and lower bound by denoting:

2³ < 10 < 2⁴

i dont exactly see how that helps. how can you find logarithms without using a calculator?

 

[Firas]

i dont exactly see how that helps. how can you find logarithms without using a calculator?

eg. log10 2^230 = 230 x log10 2
= ???

how do i find log10 2 without using a calculator?

 

[Deleted member 4993]

You can provide upper bound and lower bound by denoting:

2³ < 10 < 2⁴

From above you can say:

log₁₀(2) < (1/3)

log₁₀(2) > (1/4)

 

[Firas]

From above you can say:

log₁₀(2) < (1/3)

log₁₀(2) > (1/4)

I said i wanted an exact / quite accurate answer. how am i supposed to solve the question with those clues?

 

[Dr.Peterson]

I said i wanted an exact / quite accurate answer. how am i supposed to solve the question with those clues?

It is quite common to just estimate a number of digits based on the fact that 2^10 is quite close to 10^3. (Equivalently, log(2) is about 0.3.) This is a much better estimate than has been suggested to you. (I had intended to give this answer, but misread what others had said, and thought they had already done so.)

For a quick estimate, I would note that 2^130 = (2^10)^13 ≈ (10^3)^13 = 10^39, so it has about 40 digits. For an exact answer, I would use log(2^130) = 130 log(2) = 130*0.30103 = 39.1, so it has, in fact, 40 digits.

Note that the estimate was actually as accurate as we needed, because the estimate that log(2) = 0.3 is quite good. If you need more accuracy, you can just memorize 0.30103, which is rather nice. That will take you to very large number of digits.

 

[mmm4444bot]

… find the number of digits in an e͏xponent …

… For example, how many digits are there in 2^130 …

Here's a quick note about terminology. :cool:

2^130 = 1361129467683753853853498429727072845824

The expression on the lefthand side represents a power of two (i.e., an exponential expression), and its numerical value is shown on the right.

2 is called the "base"

130 is called the "exponent"

You're asking about finding the number of digits in the power, not the exponent. (A power is a product; the exponent indicates the number of factors in that product.)

I said i wanted an exact / quite accurate answer …

I don't find that statement here. Are you thinking of another thread?

 

[Deleted member 4993]

From above you can say:

log₁₀(2) < (1/3)

log₁₀(2) > (1/4)

log₁₀(2) ~ 1/2(.33+.25) ~ 0.29 ~ 0.3 ...... continue......

 

[JeffM]

I said i wanted an exact / quite accurate answer. how am i supposed to solve the question with those clues?

An "exact" answer? Start multiplying (I know you claim not to be able to use a calculator, but you know how to multiply without one, do you not?)

"Quite accurate answer" is a meaningless qualification. Accurate to within 1, 1%, or what?

Instead of memorizing formulas, thinking is always a good plan.

$\displaystyle 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024 \implies 2^{10} = 1024 \approx 1000 = 10^3.$

$\displaystyle 2^{230} = 2^{10} * 2^{10} * 2^{210} = 2^{10} * 2^{10} * (2^{10})^{21} \approx$

$\displaystyle 10^3 * 10^3 * (10^3)^{21} = 10^6 * 10^{63} = 10^{69}.$

$\displaystyle \therefore 2^{230} \text { is represented by about 70 decimal digits.}$

Now, as denis might correctly say, you can get an exact answer to this problem (and many similar problems) with a simple looper program. Thus, if exact answers are of practical importance, use the tools that allow you to get exact answers.

 

[Dr.Peterson]

I don't find that statement here. Are you thinking of another thread?

An "exact" answer? Start multiplying (I know you claim not to be able to use a calculator, but you know how to multiply without one, do you not?)

"Quite accurate answer" is a meaningless qualification. Accurate to within 1, 1%, or what?

I almost said the same thing, but when I looked back I realized that "find the number of digits" means, clearly, "find the exact (integer) number of digits in the number". If the number had thousands of digits, a rough estimate would be appropriate, but the 2^10=10^3 approximation makes an "exact" answer easy for reasonable numbers, since you are, after all, rounding the log up to the nearest integer. Integer answers can easily be exact. And, of course, there is no need to have an exact value for the power, in order to find the number of digits.

 

[Dr.Peterson]

log₁₀(2) ~ 1/2(.33+.25) ~ 0.29 ~ 0.3 ...... continue......

Yes, you'll get the right answer this way, but you'll be surprised that it's as accurate as it is after this sort of rough approximation. The 2^10 = 10^3 --> log(2) = 3/10 approach is much more convincing.

 

[JeffM]

I almost said the same thing, but when I looked back I realized that "find the number of digits" means, clearly, "find the exact (integer) number of digits in the number". If the number had thousands of digits, a rough estimate would be appropriate, but the 2^10=10^3 approximation makes an "exact" answer easy for reasonable numbers, since you are, after all, rounding the log up to the nearest integer. Integer answers can easily be exact. And, of course, there is no need to have an exact value for the power, in order to find the number of digits.

I agree with one small caveat. The 2^10 approximates 10^3 has a sizable relative error, in excess of 2%. Without doing a formal analysis on propagation of error, I'd hesitate before claiming an exact answer on any computation with a relative error of that magnitude even with the extra knowledge that the answer must be an integer.

SK's idea of bounds avoids the question of propagation of error.

$\displaystyle 2^3 < 10 \text { and } 2^{10} > 10^3.$

$\displaystyle \therefore 2^{230} = 2^{20} * 2^{210} = (2^10)^2 * (2^10)^21 > (10^3)^2 * (10^3)^{21} \implies$

$\displaystyle 2^{230} > 10^69.$

Therefore there are at least 70 meaningful digits in the decimal representation of 2^230.

$\displaystyle 2^{230} = 2^2 * 2^{228} = 4 * (2^3)^{71} < 4 * 10^{71}.$

Therefore there are at most 72 meaningful digits in the decimal representation of 2^230.

You can perhaps do better without a calculator, computer, or table of logarithms, but you can get the answer of 71 digits plus or minus 1 based on just arithmetic.

But as you say, you can do much better with logarithms. Except in very unusual cases where the log is very close to being an integer, the order of magnitude can be calculated exactly.

$\displaystyle y = 2^{230} \implies log_{10}(y) = 230 * log_{10}(2) \approx 69.2 << 70.$

Therefore, the exact answer is 70 digits.

I do wonder, however, what is trying to be taught with these problems. Order of magnitude is important, but the two low order digits of 2017^2017 seem to be inconsequential.

 

[Dr.Peterson]

I do wonder, however, what is trying to be taught with these problems. Order of magnitude is important, but the two low order digits of 2017^2017 seem to be inconsequential.

Good question. Obviously they would be useful in very different contexts; the most I can see is that they are teaching (if they are meant to teach at all) that we can work out facts about large numbers that involve both ends of the number (where the leftmost digit is, and what the rightmost digits are), using very different techniques. Or maybe they are just meant to test, rather than teach, the breadth of a student's knowledge.

Firas: Can you tell us the context of the questions you've been asking? Are they for a course, or for some contest or entrance exam, or what?

 

[Firas]

Good question. Obviously they would be useful in very different contexts; the most I can see is that they are teaching (if they are meant to teach at all) that we can work out facts about large numbers that involve both ends of the number (where the leftmost digit is, and what the rightmost digits are), using very different techniques. Or maybe they are just meant to test, rather than teach, the breadth of a student's knowledge.

Firas: Can you tell us the context of the questions you've been asking? Are they for a course, or for some contest or entrance exam, or what?

a contest. i may be tested on such questions so I need to know how to calculate logarithms without a calculator.

log 10 2^130 = 130 x log 10 2
= 130 x 0.3 = 39
Ans : 39 + 1 = 40

Thanks so much. However, are there any other useful results I can use?

 

[Dr.Peterson]

a contest. i may be tested on such questions so I need to know how to calculate logarithms without a calculator.

log 10 2^130 = 130 x log 10 2
= 130 x 0.3 = 39
Ans : 39 + 1 = 40

Thanks so much. However, are there any other useful results I can use?

I don't know what else they might expect you to do. Perhaps you can show us the information you are provided for this contest, so we can see what sorts of preparation would be useful.

 

[Firas]

I don't know what else they might expect you to do. Perhaps you can show us the information you are provided for this contest, so we can see what sorts of preparation would be useful.

not much info on the topics but there is practice questions. i can do most except the last few :

https://ijmo.asia/wp-content/uploads/2018/05/s3-IJMO.pdf