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Not sure how to solve this problem without graphing

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the solution(s) when: 4+x = 4(2)^x

    2. Relevant equations


    3. The attempt at a solution

    I only know how to solve this problem using graphing. I'm not sure how to do it algebraically. Please help.

    upload_2017-5-16_0-48-52.png
     
  2. jcsd
  3. May 16, 2017 #2
    I suggest that you use the fact that,$$2^x=e^{xln\left (2\right )}$$
    and expand the exponential.
     
  4. May 16, 2017 #3

    scottdave

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    Some problems do not have a nice easy algebraic solution. The (0,4) is easy enough to guess and also verify.
     
  5. May 16, 2017 #4

    Ray Vickson

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    As scottdave has indicated, some problems do no have nice solutions, and this is one of them. However, some have not-so-nice solutions, as does this one:
    $$x = 0 \; \text{or} \; x = -4 - \frac{1}{\ln 2} \text{LambertW} \left( -\frac{1}{4} \ln 2 \right) \doteq -3.690093068 $$
    Here, LambertW is a non-elementary function that does not have an explicit, finite formula, but has known series expansions and can be approximated numerically to as much accuracy as you want. You will not find it in a spreadsheet, but it is available in computer algebra systems such as Maple and Mathematica.
     
  6. May 16, 2017 #5
    Can "Lambert W function" give ##x= 0## as the solution ? without guessing ?
     
  7. May 16, 2017 #6

    ehild

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    You can apply some numerical methods. One is an iterative method (http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx)
    You should write the equation to be solved in the form x=f(x). It works in the domain where |f '(x)|<1
    The problem becomes very simple if you switch to the variable y = 2+x. The equation becomes y=2y-2. It converges quite fast for y<0.
    Substitute some initial value for y, you get the next value, y1. Substitute y1, you get the next approximation, y2.
    Continue till |yn+1-yn| are close enough.
    Starting from y= 0, we get -1, -1.5, -1.646, -1.680, -1.688, -1.689, -1.690...which corresponds to x=-3.690.
    The derivative of 2y-2= 2ylog(2) is greater then 1 for y>0.529. To get the other root (y=2), an other iterative function is needed: y=log(y+2)/log(2).
     
    Last edited: May 16, 2017
  8. May 16, 2017 #7

    Ray Vickson

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    Yes. The other solution given by Maple is
    $$x = -4 = \frac{1}{\ln 2} \text{LambertW}\left( -1, -\frac{1}{4} \ln 2 \right), $$
    where ##\text{LambertW}(-1,z)## is one of the non-analytical branches of the Lambert function, chosen so that it is real-valued on the real interval ##(-e^{-1}+i 0, 0 + i 0)## in the complex ##z##-plane. It is not easy to see, but Maple evaluates this as ##x = 0## exactly. Of course, a person would see ##x=0## right away, without use of any advanced tools.
     
  9. May 16, 2017 #8
    Wow thank you so much guys. :)
     
  10. May 16, 2017 #9

    scottdave

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    The article in lamar.edu posted by @IntegralDerivative is a nice one, explaining numerical methods.
     
  11. May 16, 2017 #10

    SammyS

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    Actually that was posted by my friend, @ehild .
     
  12. May 16, 2017 #11

    ehild

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    Thank you friend:smile:
     
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