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Math Task

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    2^n+6/5^4+n - 2 = 23

    The equation is correct. You need to work the left side to come down to the 23 = 23 conclusion. What makes this task difficult for me is the fact that on the down side of the fraction, the number is 54+n which makes it impossible to work with the upper 2n+6 on the top of the fraction. By the way, the "- 2" is not on the down side of the fraction.


    2. Relevant equations
    None.


    3. The attempt at a solution
    2^n+6/5^4+n - 2 = 23
    = 2^n+4 * 2^2/5^4+n - 2 = 23
    = (2/5)^n+4 * 4 - 2 = 23

    Couldn't go any further.
     
  2. jcsd
  3. Oct 2, 2013 #2

    SteamKing

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    FYI: 'up side' = numerator; 'down side' = denominator This is the proper terminology in dealing with fractions or other rational expressions.

    Use parentheses to clarify your expressions. I take it that the equation is:

    [itex]\frac{2^{n}+6}{5^{4}+n}[/itex]-2 = 23

    You can simplify the constants, and bring the 2 to the RHS:

    [itex]\frac{2^{n}+6}{625+n}[/itex] = 25

    After this, you can multiply both sides of the equation by the numerator (625+n):

    [itex]{2^{n}+6} = 25(625+n)[/itex]

    At this point, you can plot each side of the equation for different values of n. The intersection of the two curves will identify the approximate value of n. On the other hand, you can plug different values of n into the equation and iterate for a solution.
     
  4. Oct 2, 2013 #3

    ehild

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    Do you wanted to write [itex]2^{n}+6/5^4 + n-2 =23 [/itex] really?

    And what is the question?

    ehild
     
  5. Oct 2, 2013 #4

    haruspex

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    What you have written is (2^n) + (6/(5^4)) + n - 2 = 23. If that is not what was meant then the lack of parentheses leaves many possible interpretations, but none that I can think of lead to your second line above.
     
  6. Oct 2, 2013 #5
    No. the n+6 is together upon the 2, as well as 4+n
     
  7. Oct 2, 2013 #6

    Office_Shredder

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    So it's
    [tex] \frac{ 2^{n+6}}{5^{n+4}}-2 = 23[/tex]?

    If not you should just type it out again with much better parentheses because this guessing game can go all week.
     
  8. Oct 2, 2013 #7

    Ray Vickson

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    Use parentheses; that is what they are intended for! If you mean
    [tex]\frac{2^{n+6}}{5^{n+4}} - 2 = 23, [/tex]
    then you must write it as (2^(n+6)/5^(n+4))-2 = 23, or you could get away with 2^(n+6)/5^(n+4) - 2 = 23, because when read using standard rules the two expressions mean the same thing. It would be a bit better to leave some spaces, like this: 2^(n+6) / 5^(n+4) - 2 = 23. What YOU wrote means
    [tex] 2^n + \frac{6}{5^{n+4}}- 2 = 23 \; \text{ yes, really!}[/tex]
     
  9. Oct 2, 2013 #8
    Okay, I'm new here, and I'm not familiar with the computer math writing. So can you help me?
     
  10. Oct 2, 2013 #9

    Mark44

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    Before we can help you, we need to know exactly what the problem is. Is the problem the that's shown in posts 6 and 7?

    This is from what you posted as you started to show your work:
    Since n + 6 and 4 + n are exponents, they need parentheses to indicate that.

    2^(n+6)/5^(4+n) - 2 = 23

    The above is clear, as long as you don't mean for the -2 to be in the denominator of the fraction. If it SHOULD go in the denominator, the above should be written like this.
    2^(n+6)/(5^(4+n) - 2) = 23

    So what exactly is the problem you're working?
     
  11. Oct 2, 2013 #10
    2^(n+6)/5^(4+n) - 2 = 23
     
  12. Oct 2, 2013 #11

    Mark44

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    First thing is to add 2 to both sides.

    Next (and you did this), recognize that 2n+6 = 22 * 2n+4.
     
  13. Oct 2, 2013 #12
    Then we win

    (2^(n+4) * 2^2) / (5^(n+4)) = 25
    = (2/5)^(n+4) * 4 = 25
    = (2/5)^(n+4) = 25/4
    = (2/5)^(n+4) = (5/2)^2

    Now we can do:

    n+4 = 2
    n = 2-4
    n = -2

    From where:

    (2^(-2)) / (5^(-2)) = (5^2)/(2^2)
    = 1/(2^2) * 5^2 = (5^2)/(2^2)
    = (5^2) / (2^2) = 25/4
    = 25/4 = 25/4

    Seems right to me.
     
  14. Oct 2, 2013 #13

    Mark44

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    The above is not right, starting with the equation n+4 = 2. There's no justification for saying that if (2/5)^(n+4) = (5/2)^2, then n + 4 = 2.

    If n = -2, you have (2/5)2 = (5/2)2, which is not true.

    What I think you are forgetting is that 5/2 = (2/5)-1.

    One other thing. Don't start a line with an equation with = as you have in these three lines.
    = (2/5)^(n+4) * 4 = 25
    = (2/5)^(n+4) = 25/4
    = (2/5)^(n+4) = (5/2)^2

    An equation is not "equal" to another equation. One equation can imply another equation (==>) or be equivalent to another equation (<==>), but equations aren't equal to each other.
     
  15. Oct 2, 2013 #14
    Then solve the task if you may? You're showing me what's wrong but not how to do it right.
     
  16. Oct 2, 2013 #15

    Mark44

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    No, I may not, as PF rules don't allow finishing your work for you.
    I gave you a hint - you have 2/5 raised to a power on one side of the equation, and 5/2 raised to a power on the other side. You need to have the same base on both sides of the equation.
     
  17. Oct 2, 2013 #16
    So we have:

    (2/5)^n+4 = (2/5)^(-2)

    From where

    n+4=-2
    n=-6

    From where

    (2/5)^(-6+4) = (2/5)^(-2)
    (2/5)^(-2) = (2/5)^(-2)
     
  18. Oct 2, 2013 #17

    Mark44

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    Yes.

    You're getting better with the parentheses, but in this equation you need two more:
    (2/5)^n+4 = (2/5)^(-2)

    like this:
    (2/5)^(n+4) = (2/5)^(-2)

    Without those parentheses around n+4, what you wrote would be interpreted as this:
    (2/5)n + 4
     
  19. Oct 2, 2013 #18

    462chevelle

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    can I ask how you imply that this is a curve out of curiosity?
     
  20. Oct 2, 2013 #19

    Mark44

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    Not a curve, but two curves - y = 2n + 6 and y = 26(625 + n). The first is an exponential function; the second is a line.

    In any case, there's no point in exploring this further, as SteamKing's suggestion was based on one interpretation of what the OP meant in his first post, which was very unclear.
     
  21. Oct 2, 2013 #20

    Ray Vickson

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    Maple finds two roots:
    -(LambertW(-1/54388530464436950905813832350972787438550335255248068935623079751721324529751269656490240231959478852494267339391641703971489724175637221315534845825698544839048322133544265628848960307200*ln(2))+625*ln(2))/ln(2)
    and
    -(LambertW(-1,-1/54388530464436950905813832350972787438550335255248068935623079751721324529751269656490240231959478852494267339391641703971489724175637221315534845825698544839048322133544265628848960307200*ln(2))+625*ln(2))/ln(2)

    Both involve the Lambert W-function.
     
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