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Math Help - Logarithm square root problem

  1. #1
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    Logarithm square root problem

    Need some help with a log equation. I've done some searching and I found a very similar problem but didn't understand the response / solution.

    \sqrt{\log{x}}=\log{\sqrt{x}}

    This was the partial solution:






    set each factor equal to 0 and solve for x

    I get the first two lines, but I don't see how we go from
    to

    and I definitely don't get where to go from there.
    Note, my equation uses base 10 logs, where the solution I found uses natural logs. That shouldn't matter, though, as I'm more interested in the mechanics of how to solve it than the actual solution.

    Any help is greatly appreciated...

    Thanks in advance!
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  2. #2
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    the transition from the first two lines onwards is

     \sqrt{lnx} * \sqrt {lnx} = lnx

    because:

     (lnx)^{0.5}*(lnx)^{0.5} = lnx^{0.5+0.5} = lnx
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  3. #3
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    Dear daparsons,

    Consider,


    If you take squareroot(lnx) outside by using brackets you would get,



    It is simple. However if you have any further queries please reply.
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  4. #4
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    Quote Originally Posted by daparsons View Post
    Need some help with a log equation. I've done some searching and I found a very similar problem but didn't understand the response / solution.

    \sqrt{\log{x}}=\log{\sqrt{x}}

    This was the partial solution:






    set each factor equal to 0 and solve for x

    I get the first two lines, but I don't see how we go from
    to

    and I definitely don't get where to go from there.
    Note, my equation uses base 10 logs, where the solution I found uses natural logs. That shouldn't matter, though, as I'm more interested in the mechanics of how to solve it than the actual solution.

    Any help is greatly appreciated...

    Thanks in advance!
    Remember that \sqrt{ln(x)} =0 has no real (or complex to my knowledge) solutions
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  5. #5
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    Thanks, guys. I got it with a little substitution:

    \sqrt{\log{x}}=\log{\sqrt{x}}
    \sqrt{u}=\frac{1}{2}u
    2\sqrt{u}=u
    4=u
    \log{x}=4
    x=10,000


    Remember that has no real (or complex to my knowledge) solutions
    Yeah, that's part of what was throwing me off. Now I understand how they got to the third line, but I still don't understand how to solve for x using that process.
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  6. #6
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    Hello daparsons
    Quote Originally Posted by daparsons View Post
    ... I still don't understand how to solve for x using that process.
    The principle is this:

    • If the product of two (or more) expressions is equal to zero, then one (or more) of these expressions is itself equal to zero.

    The most common example of this is the quadratic equation, where we factorise the quadratic expression (if possible), and then put each factor in turn equal to zero. For example:
    x^2-5x+6=0

    \Rightarrow (x-2)(x-3)=0
    and it's when we put each of these factors in turn equal to zero:
    (x-2)=0
    or
    (x-3)=0
    that we get the solutions:
    x=2
    or
    x=3
    Well, we just apply the same principle here. When you get to:
    0 = \sqrt{\log x}(\tfrac12\sqrt{\log x} -1)
    you can then conclude that:
    \sqrt{\log x} = 0
    or
    (\tfrac12\sqrt{\log x} -1) = 0
    As has already been pointed out, the first of these possibilities, \sqrt{\log x} = 0, doesn't yield any values of x. So we're left with the conclusion that the only solution is when \tfrac12\sqrt{\log x}-1 = 0; i.e. x = 10,000.

    Grandad
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