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Math Help - exponential equation (With pictures!)

  1. #1
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    exponential equation (With pictures!)

    This should just about sum it up.




    Thank you interent people!
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  2. #2
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    Quote Originally Posted by Vamz View Post
    This should just about sum it up.




    Thank you interent people!
    f(x) = -2^{x - 1} + \frac{7}{2}.

    To find the x intercept, let f(x) = 0.


    So -2^{x - 1} + \frac{7}{2} = 0

    2^{x - 1} = \frac{7}{2}

    2\cdot 2^{x - 1} = 7

    2^{x} = 7

    \ln{(2^x)} = \ln{7}

    x\ln{2} = \ln{7}

    x = \frac{\ln{7}}{\ln{2}}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    f(x) = -2^{x - 1} + \frac{7}{2}.

    To find the x intercept, let f(x) = 0.


    So -2^{x - 1} + \frac{7}{2} = 0

    2^{x - 1} = \frac{7}{2}

    2\cdot 2^{x - 1} = 7

    2^{x} = 7 what happeend to ^(x-1) .. and that other 2?

    \ln{(2^x)} = \ln{7}

    x\ln{2} = \ln{7}

    x = \frac{\ln{7}}{\ln{2}}.

    Your final answer does not equal to ln(7)

    what I posted was what my teacher had in the answer key. I am assuming he's right... unless he was wrong!?
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  4. #4
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    You should know that

    a^m \cdot a^n = a^{m + n}.


    Here you have

    2^1 \cdot 2^{x - 1} = 2^{1 + x - 1} = 2^x.


    And your teacher is wrong.

    Either s/he meant \frac{\ln{7}}{\ln{2}} or s/he meant \log_2{7}.
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    Thank you.

    You sir, are a smart cookie. And I'd give you one, but I wouldn't know where to begin contacting you.


    case closed.
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  6. #6
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    Quote Originally Posted by Vamz View Post
    Thank you.

    You sir, are a smart cookie. And I'd give you one, but ...
    Hallo,

    a personal remark: Before you start throwing cookies around you could instead press the \boxed{\text{Thanks}}-button. That's the kind of food we are living on here .

    EB
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