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Thread: neg. exponent?

  1. #1
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    neg. exponent?

    Hi, I have an neg. exponent in my equation below. This is what the question asks for:

    7. Solve each equation. Check for extraneous roots.

    $\displaystyle 2^x + 12(2)^{-x} = 7$

    How would I solve this?
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  2. #2
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    $\displaystyle a^{-b} = \frac{1}{a^b} $
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  3. #3
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    that doesn't work for $\displaystyle 2^x + 12(2)^{-x} = 7$. You need to get an $\displaystyle a^2$ in it to solve this equation. My teacher was telling a person in my class yesterday to mulitply it by another a value, $\displaystyle 2^x$ in this case, to cancel out the neg. exponent.
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  4. #4
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    Quote Originally Posted by Barthayn View Post
    7. Solve each equation. Check for extraneous roots.
    $\displaystyle 2^x + 12(2)^{-x} = 7$
    How would I solve this?
    $\displaystyle 2^x + 12(2)^{-x} = 7$ is the same as $\displaystyle 2^{2x} -7\cdot 2^x+ 12=0$.

    Can you solve $\displaystyle z^2-7z+12=0?$

    The let $\displaystyle z=2^x$
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  5. #5
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    can you explain to my how did you canceled out the neg. exponent?
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  6. #6
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    multiply the equation through by $\displaystyle 2^x$ like this $\displaystyle 2^x(2^x + 12(2)^{-x} = 7) \Rightarrow 2^{2x} + 12 = 7(2^x)$
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  7. #7
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    Sure it works.

    $\displaystyle
    2^x + 12(2)^{-x} = 7 \implies 2^x + 12(\frac{1}{2^x}) = 7 \implies 2^x + \frac{12}{2^x} = 7
    $.

    Seeing it this way should scream multiply both sides by the common denominator, which is $\displaystyle 2^x $. So,

    $\displaystyle 2^x (2^x)+ \frac{12}{2^x}(2^x) = 7 (2^x) \implies
    2^{2x} + 12 = 7(2^x) $

    Now, make life easier by using the assignment $\displaystyle y=2^x$.

    The equation then becomes $\displaystyle y^2 + 12 = 7y $.

    Rewrite the equation as $\displaystyle y^2 - 7y + 12 = 0 $.

    Factoring, we get:
    $\displaystyle y=-3,-4 $.
    Last edited by mr fantastic; Dec 15th 2009 at 06:08 PM.
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  8. #8
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    Don't forget to finish the problem.


    Plugging $\displaystyle 2^x $ back in for $\displaystyle y $, we get:

    $\displaystyle
    2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)}
    $
    $\displaystyle
    2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)}
    $

    Clearly, both of these solutions are extraneous roots.
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  9. #9
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    Quote Originally Posted by abender View Post
    Don't forget to finish the problem.


    Plugging $\displaystyle 2^x $ back in for $\displaystyle y $, we get:

    $\displaystyle
    2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)}
    $
    $\displaystyle
    2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)}
    $

    Clearly, both of these solutions are extraneous roots.
    No, you are incorrect. According to my answer I got help and the answer book the answers are $\displaystyle x = 2$ and $\displaystyle x = \frac{log 3}{log 2}$. Never in my career of high school I was taught that one can put a fraction into a polynomial to make it a polynomial.

    These were the steps that one has to take to get the correct answer

    Let a = $\displaystyle 2^x$

    $\displaystyle 2^x + 12(2)^{-x} - 7 = 0$
    $\displaystyle 2a + 12-a - 7 x a = 0$
    $\displaystyle 2a^2 - 7a + 12 = 0$
    $\displaystyle (a-4)$ $\displaystyle (a-3)$
    $\displaystyle a = 4$ or $\displaystyle a = 3 $
    $\displaystyle 2^x = 4$ or $\displaystyle 2^x = 3$
    $\displaystyle x = 2$ or $\displaystyle x = \frac{log 3}{log 2}$
    Last edited by mr fantastic; Dec 15th 2009 at 06:07 PM. Reason: Trying to keep the thread nice.
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  10. #10
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    Quote Originally Posted by abender View Post
    [snip]
    Factoring, we get: Mr F says: (y - 3)(y - 4).
    $\displaystyle y=-3,-4 $. Mr F says: Therefore y = 3, 4.

    [snip]
    ..

    Quote Originally Posted by abender View Post
    Don't forget to finish the problem.


    Plugging $\displaystyle 2^x $ back in for $\displaystyle y $, we get:

    $\displaystyle
    2^x = {\color{red}3} \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{({\color{red}3})}
    $
    $\displaystyle
    2^x = {\color{red}4} \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{({\color{red}4})}
    $

    [snip]
    Minor corrections in red (you can't take the log of a negative number and get something real!!). Both solutions should now be tested in the original equation.
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  11. #11
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    The method (if not some of the details) used by abender is correct and is completely consistent with the solution given by the 'answer book'. Note:

    $\displaystyle \log_2 4 = 2$.

    $\displaystyle \log_2 3$ can be written in the form yur answer uses by using the change of base rule. Alternatively, $\displaystyle 2^x = 3 \Rightarrow \log 2^x = \log 3 \Rightarrow x \log 2 = \log 3 \Rightarrow x = \frac{\log 3}{\log 2}$.

    I will also add that the question as originally posted made no reference to a particular method, so the hint given in post #2 is completely justified.

    This thread is now closed.
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