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Math Help - help please

  1. #1
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    help please

    Given that a>0, b>0, log2^1/2 is the arithmetic mean of log4^a and log2^b. The minimum value of 2/a + 1/b?
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  2. #2
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    Re: help please

    Hello, Trefoil2727!

    First, I had to guess at what you had typed.
    Then I've made only partial progress.


    \text{Given that }a,b>0\text{ and }\log_2\tfrac{1}{2}\text{ is the arithmetic mean of }\log_4a\text{ and }\log_2b
    . . . \text{find the minimum value of }\tfrac{2}{a} + \tfrac{1}{b}

    We have: . \frac{\log_4a + \log_2b}{2} \:=\:\log_2\tfrac{1}{2} \quad\Rightarrow\quad \frac{\frac{1}{2}\log_2a + \log_2b}{2} \:=\:-1

    . . . . . . . . \tfrac{1}{2}\log_2a + \log_2b \:=\:-2 \quad\Rightarrow\quad \log_2a + 2\log_2b \:=\:-4

    . . . . . . . . \log_2a + \log_2b^2 \:=\:-4 \quad\Rightarrow\quad \log_2(ab^2) \:=\:-4

    . . . . . . . . ab^2 \:=\:2^{-4} \quad\Rightarrow\quad ab^2 \:=\:\tfrac{1}{16}

    Now what?
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  3. #3
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    Re: help please

    emm, is lg 2^(1/2), lg 4^a, lg 2^b
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  4. #4
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    Re: help please

    Hello, Trefoil2727!

    \text{Given: }\,a,b>0\text{ and }\log2^{\frac{1}{2}}\text{ is the arithmetic mean of }\log4^a\text{ and }\log2^b.
    \text{Find the minimum value of: }\,\frac{2}{a} + \frac{1}{b}

    We have: . \frac{\log4^a + \log2^b}{2} \:=\:\log2^{\frac{1}{2}} \quad\Rightarrow\quad \frac{\log\left(4^a2^b\right)}{2} \:=\:\tfrac{1}{2}\log 2

    . . . . . . . . \log\left(2^{2a}2^b\right) \;=\;\log 2 \quad\Rightarrow\quad \log\left(2^{2a+b}\right) \;=\;2^1

    . . . . . . . . 2a+b \:=\:1

    Now what?
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