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Thread: Solving equations with indeces

  1. #1
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    Solving equations with indeces

    Hello everyone,

    I am having trouble solving the very simple equation:

    $\displaystyle 2^x(2^{x-3})=32$

    I can see that $\displaystyle x=4$ is the right answer, but how could I go about showing some working for this?
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  2. #2
    Junior Member
    Joined
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    Re: Solving equations with indeces

    Quote Originally Posted by mabentley View Post
    Hello everyone,

    I am having trouble solving the very simple equation:

    $\displaystyle 2^x(2^{x-3})=32$

    I can see that $\displaystyle x=4$ is the right answer, but how could I go about showing some working for this?
    $\displaystyle 2^x(2^{x-3})=32$
    $\displaystyle 2^{x+x-3}=2^5$
    by comparing indices,
    2x-3=5
    x=4

    alternatively,
    $\displaystyle 2^x(2^{x-3})=32$
    take lg of both sides,
    $\displaystyle lg 2^{x+x-3}=lg32 $
    $\displaystyle (2x-3)lg2=lg32$
    $\displaystyle x=4$
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