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Thread: Exponential equations questions

  1. #1
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    Exponential equations questions

    Hi
    i have tried to solve these equations but still cannot get the correct answer:
    The following is what i have done:
    $\displaystyle 2^{2x}-20(2^x)=-64$
    make $\displaystyle y=2^x$
    $\displaystyle 2y^2-20y+64=0$
    $\displaystyle 2(y^2-10y+32)$
    $\displaystyle 2(y^2-10y-(\frac{10}{2})^2+32-25)$
    $\displaystyle 2(y-5)^2+7$
    $\displaystyle 2(y-5+\sqrt{7})(y-5-\sqrt{7})$
    then:
    $\displaystyle 2(y-5+\sqrt{7})$
    $\displaystyle 2y-10+2\sqrt{7}=0$
    $\displaystyle 2y+2\sqrt{7}=10$
    $\displaystyle 2y+\sqrt{7}=5$
    $\displaystyle 2y=25-7$
    $\displaystyle y=\frac{18}{2}$
    $\displaystyle 2^x=9$
    has no solution but answer says it is does.
    $\displaystyle 2(y-5-\sqrt{7})$
    $\displaystyle 2y-10-2\sqrt{7}=0$
    $\displaystyle 2y-2\sqrt{7}=10$
    $\displaystyle 2y=25+7$
    $\displaystyle y=\frac{32}{2}$
    $\displaystyle 2^x=16$
    $\displaystyle 2^x=16$
    $\displaystyle x=4$

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    $\displaystyle 2^x=9$
    has no solution but answer says it is does.
    Given the arithmetic is correct

    $\displaystyle \Rightarrow x = \log_2(9)$
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    i have tried to solve these equations but still cannot get the correct answer:
    The following is what i have done:
    $\displaystyle 2^{2x}-20(2^x)=-64$
    make $\displaystyle y=2^x$
    $\displaystyle 2y^2-20y+64=0$
    $\displaystyle 2(y^2-10y+32)$
    $\displaystyle 2(y^2-10y-(\frac{10}{2})^2+32-25)$
    $\displaystyle 2(y-5)^2+7$
    $\displaystyle 2(y-5+\sqrt{7})(y-5-\sqrt{7})$
    then:
    $\displaystyle 2(y-5+\sqrt{7})$
    $\displaystyle 2y-10+2\sqrt{7}=0$
    $\displaystyle 2y+2\sqrt{7}=10$
    $\displaystyle 2y+\sqrt{7}=5$
    $\displaystyle 2y=25-7$
    $\displaystyle y=\frac{18}{2}$
    $\displaystyle 2^x=9$
    has no solution but answer says it is does.
    $\displaystyle 2(y-5-\sqrt{7})$
    $\displaystyle 2y-10-2\sqrt{7}=0$
    $\displaystyle 2y-2\sqrt{7}=10$
    $\displaystyle 2y=25+7$
    $\displaystyle y=\frac{32}{2}$
    $\displaystyle 2^x=16$
    $\displaystyle 2^x=16$
    $\displaystyle x=4$

    P.S
    Your substitution should be...
    $\displaystyle y^2-20y+64=0$

    Not

    $\displaystyle 2y^2...$
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  4. #4
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    " 2y^2-20y+64=0 " is incorrect.

    2^(2x)-20(2^x)=-64
    " Let 2^x be y " is the right direction, but-

    Your mistake: 2^(2x) = 2(2^x)^2.

    2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

    As for the rest of the steps, I believe you know them already
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    $\displaystyle 2^{2x}-20(2^x)=-64$
    Yikes! What are you doing? Keep it simple:

    2^2(2^x) - 20(2^x) = -64
    4(2^x) - 20(2^x) = -64
    -16(2^x) = -64
    2^x = -64 / -16
    2^x = 4
    x = 2
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  6. #6
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    lols, Wilmer.

    2^2(2^x) = 2^(x+2) is not the same as 2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x)
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  7. #7
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    Quote Originally Posted by werepurple View Post
    " 2y^2-20y+64=0 " is incorrect.

    2^(2x)-20(2^x)=-64
    " Let 2^x be y " is the right direction, but-

    Your mistake: 2^(2x) = 2(2^x)^2.

    2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

    As for the rest of the steps, I believe you know them already
    oh oops, i think i'll tired it again and i should be able to get it.
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  8. #8
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    lol, i didn't ever need to complete the square and i answer my own question.
    i just made it:
    $\displaystyle (y-16)(y-4)$
    which will therefore get me x=4 and x=2.

    oh well, thanks for spotting the mistake
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  9. #9
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    Quote Originally Posted by werepurple View Post
    lols, Wilmer.
    2^2(2^x) = 2^(x+2) is not the same as 2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x)
    right...BUT tell me why I got the correct answer
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  10. #10
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    Sheer coincidence, Wilmer. Haha, and that is why there are intersections of lines, curves, and etc; values coincide. ^^
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