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Another logarithm question.

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    by substituting y = log2x solve for x in the following equation:

    √log2x = logs2√x
    2. Relevant equations

    logab=c then a^c = b
    3. The attempt at a solution

    if y = log2x then the equation becomes √y = log2 x^1/2
    this implies √y = 1/2 log2x which simplifies to √y = 1/2 y
    [√y]^2 = [ 1/2 y]^2
    y = (y^2)/4
    4y = y^2
    4y-y^2 = 0
    y(4-y) = 0
    4-y = 0
    y = 4
    if y = 4 and y = log2x then 4 = log2x
    if loga b = c then a ^c = b
    this implies that 2^4 = x and x = 16. anyone agrees with this solution
     
  2. jcsd
  3. Oct 22, 2011 #2

    SammyS

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    Is this the equation you're supposed to be solving
    [itex]\sqrt{\log_2\,x\ }=\log_2\,\sqrt{x}\ \ ?[/itex]​

    The equation 2u2 = u , has two solutions. So does the equation [itex]2y=\sqrt{y}\,.[/itex]

    Write 2u2 = u as 2u2 - u = 0, then factor out the common factor.
     
  4. Oct 22, 2011 #3

    HallsofIvy

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    SammyS, isn't that exactly what he said he did?

    Doublell, it's easy to check your answer. If x= 16 then [itex]\sqrt{x}= 4[/itex] and [itex]log_2(\sqrt{x})= log_2(4)= log_2(2^2)= 2[/itex]. Of course, [itex]log_2(16)= log_2(2^4)= 4[/itex] so [itex]\sqrt{log_2(x})= \sqrt{4}= 2[/itex] also.
     
  5. Oct 22, 2011 #4

    SammyS

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    Well, I admit that I didn't read his post as carefully as I should have. (I may have spent too much time working with another PH user, and some of his behaviors were contagious.) However, what I should have pointed out, is that if y(4 - y) = 0, there are two solutions for y. OP did drop the y = 0 solution.

    If log2(x) = 0, then x = 1.
     
  6. Oct 22, 2011 #5
    Writing my posts more clearly

    i noticed that my post are not as clear as u guys eg i write log2x when in ur posts its clear to understand any advice on how i can post my questions in a similar fashion as yours?
     
  7. Oct 22, 2011 #6

    Mark44

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    Re: Writing my posts more clearly

    Please - no textspeak (e.g., u and ur). Using textspeak is a violation of forum rules.

    You can write exponents and subscripts using the expanded menu that is available when you click Go Advanced. For subscripts, as in log2(x), click the X2 button and enter the subscript. (It doesn't have to be 2.)

    For exponents, as in w4, click the X2 button and enter the exponent. There are a bunch of other symbols that you can use, shown to the right of the text-entry window, such as √, ≤, Ʃ, ±, and Greek letters.
     
  8. Oct 22, 2011 #7
    Re: Writing my posts more clearly

    thanks and i will remember no text speaking
     
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