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Thread: Logarithm

  1. #1
    Member classicstrings's Avatar
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    Logarithm

    How would I solve this?

    2^2x - 2^(x+1) = 0

    Having trouble with this too

    2^2x - 12 * 2^x + 32 = 0
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  2. #2
    Member Glaysher's Avatar
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    $\displaystyle 2^{2x}=2^{x+1}$

    $\displaystyle 2x=x+1$

    $\displaystyle x=1$
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  3. #3
    Member Glaysher's Avatar
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    For second one

    $\displaystyle (2^x - 8)(2^x - 4) = 0$

    $\displaystyle 2^x =8$ gives $\displaystyle x=3$

    $\displaystyle 2^x =4$ gives $\displaystyle x=2$
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  4. #4
    Member classicstrings's Avatar
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    Thanks Glaysher!
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