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Solve for x

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    5^2x - 4(4^x) + 8 = 5

    2. Relevant equations



    3. The attempt at a solution

    5^2x - 16^x + 8 = 5

    5^2x - 16^x + 8 - 5 = 0
     
  2. jcsd
  3. Jul 7, 2011 #2

    SammyS

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    4(4x) = 41 4x = 4(x+1). (This is not the same as 16x.)

    I suggest that you review properties of exponents.

    It's difficult to work with logarithms (which are actually exponents) if you don't know how to work with exponents.
     
  4. Jul 7, 2011 #3
    i dont properties of exponents
     
  5. Jul 7, 2011 #4

    I like Serena

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    Let me check this.

    You write:
    5^2x - 4(4^x) + 8 = 5 ​


    So did you mean:
    [tex]5^2x - 4(4^x) + 8 = 5[/tex]
    or:
    [tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
    or yet something else?

    Actually, I can't imagine you meant either the first or the second, because you won't be able to solve either.
    You would only be able to solve the second form by trial and error, but I suspect that was not intended.
     
  6. Jul 7, 2011 #5
    [tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
     
  7. Jul 7, 2011 #6

    HallsofIvy

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    On the other hand, if it were
    [tex]4^{2x} - 4(4^x) + 8 = 5[/tex]
    It would be easy- using the "laws of exponents". If you do not know the laws of exponents ([itex]a^x*a^y= a^{x+ y}[/itex], [itex](a^x)^y= (a^y)^x= a^{xy}[/itex]), you should not be attempting a problem like this. Who ever gave you this problem clearly believes that you do know them. Learn the laws of exponents!
     
  8. Jul 7, 2011 #7

    Ray Vickson

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    An obvious solution is x = 0. Numerical solution methods home in on this solution as well.

    RGV
     
  9. Jul 7, 2011 #8
    yes i am aware of those two laws but how will i apply it to:
    [tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
     
  10. Jul 7, 2011 #9

    I like Serena

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    You won't.
    You'll only solve it numerically, but I do not think your current course is teaching you that.
    (I still think you made a copying error when you typed in the problem. :wink:)
     
  11. Jul 7, 2011 #10
    oh ok
     
  12. Jul 7, 2011 #11
    no i did not make any error that is what i am seeing on the paper
     
  13. Jul 7, 2011 #12

    I like Serena

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    Then the only way you'll solve it, is by trial and error (aka numerically).

    Try filling in x=0, x=1, x=-1, x=2.
    Make a graph.
    Try and think of other values for x to try, like x=0.5.

    There! :smile:
     
  14. Jul 7, 2011 #13
    ok thanks
     
  15. Jul 7, 2011 #14

    SammyS

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    You can solve it graphically.

    [itex]5^{2x} - 4(4^x) + 8 = 5[/itex] is equivalent to [itex]25^{x} - 4(4^x) + 3 = 0[/itex]

    Graph [itex]y= 25^{x} - 4(4^x) + 3[/itex] and find the y-intercepts.
     
  16. Jul 7, 2011 #15
    ok i will try that
     
  17. Jul 7, 2011 #16

    HallsofIvy

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    I will suggest, once again, that you look at the problem again and make sure it is not
    [tex]4^{2x}−4(4^x)+8=5[/tex]
    which, as I said before, would be easy.
     
  18. Jul 7, 2011 #17
    yep, its not 4 it is 5
     
  19. Jul 7, 2011 #18

    SammyS

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    Hello, HoI, (Check for a PM)

    While you are correct that [itex]4^{2x}−4(4^x)+8=5[/itex] would be a more reasonable problem to solve, the solutions to [itex]5^{2x}−4(4^x)+8=5[/itex] are rational, which is surprising to me.
     
    Last edited: Jul 7, 2011
  20. Jul 7, 2011 #19
    ok then, thanks
     
  21. Jul 7, 2011 #20
    I'm not sure if this works, and it's definitely beyond the scope of precalculus, but we could try to use the generalized binomial theorem to look for rational solutions at least:

    [tex]
    \begin{align*}
    0 &= 5^{2x} - 4 \cdot 4^x + 3 = (1+4)^{2x} - 4^{x+1} + 3 = \sum_{k=0}^\infty \binom{2x}{k} 4^k - 4^{x+1} + 3 = \sum_{k=1}^\infty \binom{2x}{k} 4^k - 4^{x+1} + 4 \\
    &= 4 \left(\sum_{k=1}^\infty \binom{2x}{k} 4^{k-1} - 4^x + 1 \right).
    \end{align*}
    [/tex]

    (Strictly speaking the infinite series might diverge, but I think it might converge 2-adically.) But I'm not sure how to proceed now... Expand [itex] 4^x [/itex] as a power series?
     
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