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Math Help - Why does this equation hold

  1. #1
    Newbie bakerconspiracy's Avatar
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    Why does this equation hold

    Why does this equation hold true? I cannot seem to algebraically prove this.

    2^{log n} = n^{log 2}

    thanks for the help in advance.
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  2. #2
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    Re: Why does this equation hold

    Quote Originally Posted by bakerconspiracy View Post
    Why does this equation hold true? I cannot seem to algebraically prove this.

    2^{log n} = n^{log 2}

    thanks for the help in advance.
    It only holds true for the values of n such that n^2 = 2^n.
    Thanks from bakerconspiracy
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  3. #3
    Newbie bakerconspiracy's Avatar
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    Re: Why does this equation hold

    In my homework we have a solution that reduces from 2^{lg^2(n) } = 2^{lg(n) * lg(n)} = n^{lg(n)} where lg(n) = log_2(n). Can you explain how this happens then? Does it only work for the values you described?

    Quote Originally Posted by Prove It View Post
    It only holds true for the values of n such that n^2 = 2^n.
    Last edited by bakerconspiracy; June 10th 2012 at 11:36 PM.
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  4. #4
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    Re: Why does this equation hold

    Quote Originally Posted by bakerconspiracy View Post
    Why does this equation hold true? I cannot seem to algebraically prove this.

    2^{log n} = n^{log 2}

    thanks for the help in advance.
    I do apologise, I misread your question. I thought the LHS was \displaystyle \begin{align*} 2\log{(n)} \end{align*}, not \displaystyle \begin{align*} 2^{\log{(n)}} \end{align*}, and the RHS was \displaystyle \begin{align*} n\log{(2)} \end{align*}, not \displaystyle \begin{align*} n^{\log{(2)}} \end{align*}.

    Anyway, the best thing to do is to convert them to the same base

    \displaystyle \begin{align*} 2^{\log{(n)}} &\equiv e^{\log{\left[2^{\log{(n)}}\right]}} \\ &\equiv e^{\log{(n)}\log{(2)}} \\ &\equiv \left[e^{\log{(n)}}\right]^{\log{(2)}} \\ &\equiv n^{\log{(2)}} \end{align*}

    Q.E.D.
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  5. #5
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    Re: Why does this equation hold

    Hello, bakerconspiracy!

    \text{Prove: }\:2^{\log n} = n^{\log 2}

    \text{Let }X \:=\:2^{\log n}

    \text{Take logs: }\;\log X \:=\:\log\left(2^{\log n}\right) \:=\:\log n\cdot\log 2 \:=\:\log 2\cdot\log n

    . . . . . . . . \log X \;=\;\log\left(n^{\log 2}\right)

    \text{Exponentiate: }\;X \;=\;n^{\log 2}


    \text{Therefore: }\;2^{\log n} \;=\;n^{\log 2}
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