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Thread: possibly a division log problem

  1. #1
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    possibly a division log problem

    I think I might have to subtract? possibly a division log problem-log-5-q.jpg
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  2. #2
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    Re: possibly a division log problem

    please type out your problem ...
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  3. #3
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    Re: possibly a division log problem

    You think? You should already have learned the "laws of exponents":
    a^na^m= a^{n+ m}, \frac{a^n}{a^m}= a^{n- m}a, and (a^n)^m= a^{mn}.
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  4. #4
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    Re: possibly a division log problem

    Is the answer 8^2x/2^4x=4^-2x?
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    Re: possibly a division log problem

    Quote Originally Posted by dtdj13 View Post
    Is the answer 8^(2x)/2^(4x) = 4^(-2x)?
    No ...

    oh,yea ...parentheses!


    $\dfrac{8^{2x}}{2^{4x}} = \dfrac{(2^3)^{2x}}{2^{4x}} = \dfrac{2^{6x}}{2^{4x}}$

    finish it ...
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  6. #6
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    Re: possibly a division log problem

    it is a multiple choice question the answers are 2^x, 4^X, 2^-2x,4^-2x i got 2^2x which isn't one of the choices.
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    Re: possibly a division log problem

    Again using the "laws of exponents", 2^{2x}= (2^2)^x. Learn the laws of exponents!
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    Re: possibly a division log problem

    Quote Originally Posted by dtdj13 View Post
    it is a multiple-choice question the answers are 2^x, 4^X, 2^-2x,4^-2x i got 2^2x which isn't one of the choices.
    No, use parentheses (as in the following):

    ... the answers are 2^x, 4^x, 2^(-2x), 4^(-2x). I got 2^(2x), which isn't one of the choices.

    Suppose the actual answer is equivalent to 2^(2x), which is what you figured.

    Can you convert that to one of the answer choices?

    2^{2x} \ = \ 2^{2(x)} \ = \ (2^2)^x \ = \ ?
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    Re: possibly a division log problem

    Quote Originally Posted by dtdj13 View Post
    it is a multiple choice question the answers are 2^x, 4^X, 2^(-2x),4^(-2x) i got 2^(2x) which isn't one of the choices.
    parentheses ...
    parentheses ...
    parentheses ...
    parentheses ...

    btw, did I say ...

    parentheses ... ?
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  10. #10
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    Re: possibly a division log problem

    is the answer equivalent to 4^x? or is it 2^x

    btw I don't how to use parentheses outside of using them in a calculator
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  11. #11
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    Re: possibly a division log problem

    put away the calculator ...

    $(\color{red}{2^2})^x = (\color{red}{4})^x = 4^x$
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  12. #12
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    Re: possibly a division log problem

    The conventions for use of parentheses in a calculator and anywhere else are the same.

    Failure to use parentheses where they are needed will cause you to get erroneous answers.
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