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Calculating 3^0.2 without using a calculator?

  1. Jan 1, 2005 #1
    How can i calculate 3^0.2 without using a calculator?

    We can write 3^4 as 3*3*3*3. Can 3^4.2 be written in the same way (without exponent)?
    Last edited: Jan 1, 2005
  2. jcsd
  3. Jan 1, 2005 #2

    matt grime

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    no. well, not in any meangfully simlar fashion. You're asking for the fifth root of 3. This is an irrational number, you cannot write it as a product of integers or rational numbers.
  4. Jan 1, 2005 #3
    Do you mean to say that an integer with a fractional exponent cannot be written in terms of multiplication? What about in terms of multiplication, addition and subtraction? Are there any proofs concerning this?? Is there anything like 'level' of operators?
  5. Jan 1, 2005 #4

    matt grime

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    In general x^n, if n is not an integer, and x is an integer, will be irrational, and irrational numbers by definition cannot be written in the form a/b where a and b are integers.

    of course 8^{1/3}=2, so it doesn't always hold.

    If we assume that by "in terms of multiplication, addition and subtraction" you mean a finite number of operations on integers, then any such expression is a rational number, and so we have answered the question as "no there is no way of wriiting it in the way you ask", and the proof is fairly straightforward - you know how to prove the square root of 2 is irrational?

    However, there are other "infinite" operations we could perform it depends on what yoyu want to do with it.

    If you know about continued fractions then there are nice representations of lots of irrational numbers.

    tell me what you think level of operators means i can think about it.

    remember every real number has a decimal expansion, and sometimes we can compute these to as many places as you care to name.
  6. Jan 2, 2005 #5
    thanks...i'll do some reading...and post here again in about 15 hours.
  7. Jan 2, 2005 #6
    How does the calculator do it?
  8. Jan 2, 2005 #7


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    I don't know what a calculator does,but surely i'd like to tell what i'd do.
    Okay,let's call the number u wanna compute by 'N'.Then
    [tex] N=3^{0.2} [/tex]
    Take natural logarithm of both sides.U'll get
    [tex] \ln N=0.2\ln 3 [/tex]
    Now search in a logarithm table 'ln 3',multiply the number by 0.2 (which mean division by 5),call the number by 'a' and then u have this equation:
    [tex] \ln N=a [/tex]
    Search in the same table of logarithms the N for which it's logarithm equals 'a'.
    In the past,when computers were SciFi,logarithm tables were very handy and used.

  9. Jan 2, 2005 #8
    Where did the log tables come from?
  10. Jan 2, 2005 #9
    A more elegant statement is: If [tex]a=m/n[/tex] is rational and non-integral, and [tex]x[/tex] is any integer, then if [tex]x^a[/tex] is not integral it must be irrational.


    Let [tex]x^a=x^{m/n}, x,m,n\in\mathbb{N}[/tex] be non-integral; assume for contradiction that it is rational. Then [tex]x^{m/n}=\frac{p}{q}[/tex] for some relatively prime [tex]p,q[/tex]. Thus [tex](x^{m/n})^n=x^m=(\frac{p}{q})^n[/tex]; and since [tex]p^n,q^n[/tex] are relatively prime, [tex](\frac{p}{q})^n[/tex] must be non-integral; therefore [tex]x^m[/tex] is non-integral, a contradiction.
    Therefore [tex]x^{m/n}[/tex] is irrational.
  11. Jan 2, 2005 #10


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    I'll explain why the fundamental transcendental equation is
    [tex] e^{x} =10 [/tex] (1)
    ,with the solution [itex] x=\ln 10 [/tex].
    This is the fundamental number for the construction of natural logarithm tables.

    Using [tex] \int_{1}^{10} \frac{dx}{x} =\ln 10 [/tex]
    and computing the integral using Riemann sums and very fine divisions,people were able to compute this fundamental number with great accuracy.

    The fundamemental 2 formulas which help build natural logarithm tables are
    [tex] \ln a^{b}=b\ln a [/tex]
    [tex] \ln a+\ln b =\ln (ab) [/tex]
    ,and these two need to be combined with the formula (series) that gives the natural logarithm for every real number between 1 and 2
    [tex] \ln(1+x)=\sum_{k=1}^{+\infty} (-)^{n+1}\frac{x^{n}}{n} [/tex]
    ,with [itex] x\in [0,1] [/itex].

    1)Let's compute [itex] \ln 237 [/itex]
    [tex] \ln 237=\ln 200+\ln 1.185=\ln 2+\ln 100+\ln 1.185=\ln 2+2\ln 10+\ln 1.185 [/tex]
    So,u make use of the series to find the natural logarithms of 2 and 1.185 and use the fundamental number [itex] \ln 10 [/itex].

    2)Let's compute [tex] \ln \sqrt[3.15]{\frac{\pi}{8.26}} [/tex].Take [itex]\pi=3.14 [/itex]
    [tex] \ln \sqrt[3.15]{\frac{\pi}{8.26}} =\frac{1}{3.15}(\ln \pi-\ln 8.26)=\frac{1}{3.15}(\ln 2+\ln 1.57-3\ln 2-\ln 1.0325) [/tex]

    And so on and so forth.

  12. Jan 2, 2005 #11
    Here is an interesting method I once new, but I'm not sure if it works for fifth roots. Of course, since most roots are irrational, you must stop at some point and round, but it is technically possible to find the square root of any number by hand.
  13. Jan 2, 2005 #12


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    I've been taught this method when i was in my 6-th grade.10 years ago,to be more exact.For the third order,it's rather tricky,it's not that simple,u have to find cubes and that's a bit difficult if u have large numbers,of 12,20,50 digits.So save it only for the second order and use pencil and logarithm tables for other roots,if u don't have a calculator.

  14. Jan 2, 2005 #13
  15. Jan 2, 2005 #14


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    bigyan, yr 15 hrs r up.
    A good way to do it is with Newton's method. You need to find the roots of [itex]x^5-3=0[/itex]. First you try a few fractions and find the solution is close to x_1=5/4. Then from Newton's method,
    [tex]x_{n+1}=x_n-{x_n^5-3\over 5x_n^4}[/tex]
    This gives a next approx: x_2=3893/3125, which is closer than a part in 30,000.
    Apply it again and you get 9 decimal places, but that's tedious w/o calculator.
  16. Jan 2, 2005 #15


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    what about expanding the binomial theorem with fractionl ,exponents, also due to newton?
  17. Jan 3, 2005 #16
    If you want to find 3^.2, just go to google, hehe. I would like to see the answer to this also, as earlier, I wanted to find out how to do 25^.5 without a calculator, and just to let you know, raising any number to the power if .5 gives you the square root. root=Number^1/root_number
  18. Jan 3, 2005 #17
    [tex]25^0^.^5[/tex] = [tex]25^\frac{1}{2}[/tex]

    This is because [tex]0.5[/tex] = [tex]\frac{1}{2}[/tex]

    [tex]x[/tex] to the power of [tex]\frac{1}{2}[/tex] = [tex]\sqrt{x}[/tex]

    Therefore [tex]25^\frac{1}{2}[/tex] = [tex]\sqrt{25}[/tex] = [tex]5[/tex]

    The Bob (2004 ©)
  19. Jan 3, 2005 #18


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    3.2= 31/5 so if x= 3.2, x5= 3.

    Probably the best way to solve that (without a calculator but still using the basic arithmetic operations) is to use "Newton's method". Newton's method for solving the equation f(x)= 0 is to set up a sequence of numbers xn defined by taking x0 as some initial "guess" and then xn+1= xn- f(xn/f '(xn) where f '(x) is the derivative of f.,

    Let f(x)= x5- 3. Then f '(x)= 5x4. The formula becomes
    xn+1= xn- (xn5-3)/(5xn4)= (4/5)xn+(1/5)(3/xn4).

    If we start with x0= 2, say, then x1= 1.6375. Now repeat.

    One way of thinking about this is that if x5= 3, then x= 3/x4.
    If, on the other hand, x is less than the actual fifth root of 3, 3/x4 will be larger and if x is larger, 3/x4 is less than the actual root: in either case, the actual root is between x and 3/x4 and any "weighted" average of the two will move close to the true root. Newton's method gives x a weight of 4/5 while 3/x4 has a weight of 1/5 in the average.
    Last edited: Jan 3, 2005
  20. Jan 4, 2005 #19


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    The Babylonians used to have a method that was as accurate as Newton's method, even if less efficient. Actually, if you try Newton's method and the Babylonian method for square roots, you'll notice Newton's method winds up skipping every other guess. I haven't actually tried the Babylonian method for higher roots - Newton's method is more than enough tedium when done manually. (The point being that some kind of effective method for solving such problems has been in existence for a long time).

  21. Jan 4, 2005 #20
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