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Solve for x

  1. Oct 5, 2009 #1
    log x + (log x)^2 = o

    It is very unclear to me how to solve this. I have managed to find x = 1, but cannot find x = 1/10. Also I have no idea if i am doing it right. How do i solve for x?

    By log x i mean the common log
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2

    Mark44

    Staff: Mentor

    This is a quadratic in log x. Further, there is no constant term in this quadratic, so it can be factored.
     
  4. Oct 5, 2009 #3
    log4 x2 = (log4 x) 2

    why does log4 x = 0

    i have found that the x = 16, but i am confused of why x also is 1?
     
  5. Oct 5, 2009 #4
    Where did you get log4 with base 4? If one of the answers is 1/10 as you gave in your first post, you're working with log10 with base 10. Try factoring the left hand side of the equation.
     
  6. Oct 5, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]log_4(x^2)= 2 log_4(x)[/itex] so your equation is the same as [itex]2log_4(x)= log_4(x)[/itex]. Subtracting [itex]log_4(x)[/itex] from both sides, you get [itex]log_4(x)= 0[/itex] which has x= 1 as its only solution.

    x= 16 is NOT a solution. [itex]16= 4^2[/itex] so [itex]16^2= (4^2)^2= 4^4[/itex]. [itex]log_4(16^2)= 4[/itex] while [itex]log_4(16)= 2[/itex]. They are NOT equal.

    Did you mean [itex](log_4(x))^2= log_4(x)[/itex]? You can write that as [itex](log_4(x))^2- log_4(x)= log_4(x)(log_4(x)- 1)= 0[/itex]. Then either [itex]log_4(x)= 0[/itex], with gives x= 1, or [itex]log_4(x)- 1= 0[/itex] so that [itex]log_4(x)= 1[/itex] and x= 4. But x= 16 is still not a solution.
     
    Last edited: Oct 6, 2009
  7. Oct 5, 2009 #6
    sorry to clarify it is a whole new equation. The second post helped me figure the original equation out. My fault for not saying its a new equation
     
  8. Oct 5, 2009 #7


    x = 16 is a solution of [itex]\log_4(x^2) = (\log_4 x)^2.[/itex] Both sides equal 4.

    HallsofIvy's comment though gives the hint for turning this into a (factorable) quadratic equation in log4x which has 2 solutions, namely 1 and 16.

    --Elucidus
     
  9. Oct 6, 2009 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Ah, I missed the square on the x on the left side! I need to get my eyes examined!
     
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