Sum of the Values of X, Exponential Equation

  1. 1. The problem statement, all variables and given/known data
    (4x)^(1 + log(base 2) (x)) = 8(x^3)
    What is the sum of the values of x that fullfill that equation?
    A) 2.5
    B) 2.0
    C) 1.5
    D) 1.0
    E) 0.5


    2. Relevant equations
    Use the exponential equation only and make the lower one (exponented) 1.


    3. The attempt at a solution
    (4x) = (2 x^(1/2))^2
    8(x^3) = (2x)^3
    If I insert x = (1/4) to make 4x = 1 it doesn't fulfill the equation..
    So does x = (1/2)..
    I cannot simplify the exponential equation... Any assistance please?
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
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    hi wiraimperia! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)

    for any number n, what is nlog2(x) ? :wink:
     
  4. Simon Bridge

    Simon Bridge 14,660
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    [tex]4x(4x)^{\log_2x} = 8x^3[/tex]
    ... how many values of x satisfy the equation?

    I'd recheck for x=1/2 ...
     
    Last edited: Jul 6, 2012
  5. I mean if it has 2 solutions, then we are asked X1 + X2, if it has 3, then X1 + X2 + X3, and so on...
     
  6. ehild

    ehild 11,928
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    Try to take the logarithm of both sides.

    ehild
     
  7. tiny-tim

    tiny-tim 26,054
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    for any number m, what is (2m)log2(x) ? :wink:
     
  8. Simon Bridge

    Simon Bridge 14,660
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    Is there any way you can figure out how many solutions it is likely to have?
    Have you rechecked if x=1/2 is a solution? Personally I managed it by systematic guesswork using the list of possible solutions ... but I've had practice.

    Have you tried any of the other suggestions and hints? They are all good.

    Is there a particular method you are supposed to use or can we throw anything we like at the problem? eg. there's always brute-force methods like plotting the curves to get ballpark figures and then using Newton/Raphson ...
     
  9. Taking logarithm on both sides with base 2, you get:
    [tex](1+log_2 x)log_2(4x)=3+3log_2 x[/tex]

    It is easy to solve now. :smile:
     
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