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Math Help - 4^(2log(base8)7), equivalent expression, integers

  1. #1
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    4^(2log(base8)7), equivalent expression, integers

    Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.

    Could someone point me in the right direction?
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  2. #2
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    Quote Originally Posted by jsmith90210 View Post
    Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.

    Could someone point me in the right direction?

    First, convert 2Log8(7) > Log8(49)

    Then, I believe you can use the following rule:
     <br />
a^{\text{Logb}(c)}=c^{\text{Logb}(a)}<br />
    Therefore,
     <br />
4^{\text{Log8}(49)}=49^{\text{Log8}(4)}<br />
    Log8(4) = 3/2

    So, you have
    49^(3/2)


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    Josh
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    Last edited by Invertible; February 21st 2009 at 11:26 PM. Reason: Add Tag Line
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  3. #3
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    Quote Originally Posted by Invertible View Post
    First, convert 2Log8(7) > Log8(49)

    Then, I believe you can use the following rule:
     <br />
a^{\text{Logb}(c)}=c^{\text{Logb}(a)}<br />
    Therefore,
     <br />
4^{\text{Log8}(49)}=49^{\text{Log8}(4)}<br />
    Log8(4) = 3/2 ...... Unfortunately no.

    So, you have
    49^(3/2)


    ---------------------------------
    Josh
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    ......
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  4. #4
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    Quote Originally Posted by earboth View Post
    ......
    My mistake, I believe that:
    Log8(4) = 2/3

    Therefore, the corrected answer is
    49^{(2/3)}
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  5. #5
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    what is the name of that rule?
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  6. #6
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    Hello, jsmith90210!

    A different approach . . .


    Simplify 4^{2\log_87} into the form a^{\frac{b}{c}}, all integers.

    We have: . 4^{\log_8(7^2)} \:=\:4^{\log_8(49)} .[1]

    Let \log_8(49) = p \quad\Rightarrow\quad  8^p \,=\,49\quad\Rightarrow\quad \left(2^3\right)^p \,=\,49 \quad\Rightarrow\quad 2^{3p} \,=\,49

    . . Then: . 3p \,=\,\log_2(49) \quad\Rightarrow\quad p \:=\:\tfrac{1}{3}\log_2(49)<br />

    . . Hence: . \log_8(49) \:=\:\tfrac{1}{3}\log_2(49)


    Substitute into [1]: . 4^{\frac{1}{3}\log_2(49)} \;=\;\left(2^2\right)^{\frac{1}{3}\log_2(49)}

    . . . . . =\;2^{\frac{2}{3}\log_2(49)} \;=\; 2^{\log_2(49^{\frac{2}{3}})} \;=\;\boxed{49^{\frac{2}{3}}}



    This answer is correct ... it satisfies the requirements of the problem.

    But I bet that "they" simplified it further: . 49^{\frac{2}{3}} \:=\:\left(7^2\right)^{\frac{2}{3}} \:=\:7^{\frac{4}{3}}

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