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Math Help - 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.

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

    How would I solve this?
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  2. #2
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     a^{-b} = \frac{1}{a^b}
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  3. #3
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    that doesn't work for 2^x + 12(2)^{-x} = 7. You need to get an 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, 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.
     2^x + 12(2)^{-x} = 7
    How would I solve this?
     2^x + 12(2)^{-x} = 7 is the same as  2^{2x} -7\cdot 2^x+ 12=0.

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

    The let 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 2^x like this 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.

    <br />
2^x + 12(2)^{-x} = 7 \implies 2^x + 12(\frac{1}{2^x}) = 7 \implies 2^x + \frac{12}{2^x} = 7<br />
.

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

     2^x (2^x)+ \frac{12}{2^x}(2^x) = 7 (2^x) \implies <br />
2^{2x} + 12 = 7(2^x)

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

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

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

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


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

     <br />
2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)}<br />
    <br />
2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)}<br />

    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  2^x back in for  y , we get:

     <br />
2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)}<br />
    <br />
2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)}<br />

    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 x = 2 and 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 = 2^x

    2^x + 12(2)^{-x} - 7 = 0
    2a + 12-a - 7 x a = 0
    2a^2 - 7a + 12 = 0
    (a-4) (a-3)
    a = 4 or a = 3
    2^x = 4 or 2^x = 3
    x = 2 or x = \frac{log 3}{log 2}
    Last edited by mr fantastic; December 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).
     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  2^x back in for  y , we get:

     <br />
2^x = {\color{red}3} \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{({\color{red}3})}<br />
    <br />
2^x = {\color{red}4} \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{({\color{red}4})}<br />

    [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:

    \log_2 4 = 2.

    \log_2 3 can be written in the form yur answer uses by using the change of base rule. Alternatively, 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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