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  1. #1
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    solve

    solve
    2^{\sqrt{x}}+2^x=20

    i know its 4 but how
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  2. #2
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    Quote Originally Posted by perash View Post
    solve
    2^{\sqrt{x}}+2^x=20

    i know its 4 but how
    You'd suspect a perfect square solution so you'd try trial and error ..... try x = 1, 4, 9 ..... whoops go back a second x = 4 aha!


    Alternatively:

    Let 2^{\sqrt{x}} = w. Then:

    w + w^2 = 20 \Rightarrow w^2 + w - 20 = 0 \Rightarrow (w - 4)(w + 5) = 0 \Rightarrow w = 4, \, -5.

    Therefore:

    2^{\sqrt{x}} = 4 \Rightarrow 2^{\sqrt{x}} = 2^2 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4.

    2^{\sqrt{x}} = -5 has no real solution.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You'd suspect a perfect square solution so you'd try trial and error ..... try x = 1, 4, 9 ..... whoops go back a second x = 4 aha!


    Alternatively:

    Let 2^{\sqrt{x}} = w. Then:

    w + w^2 = 20 \Rightarrow w^2 + w - 20 = 0 \Rightarrow (w - 4)(w + 5) = 0 \Rightarrow w = 4, \, -5.

    Therefore:

    2^{\sqrt{x}} = 4 \Rightarrow 2^{\sqrt{x}} = 2^2 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4.

    2^{\sqrt{x}} = -5 has no real solution.
    thank you mr fantastic
    but if
    2^{\sqrt{x}} = w
    then
    2^{2\sqrt{x}} = w^2
    not

    2^{{x}} = w^2
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  4. #4
    Moo
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    Hi,

    Is x an integer ?

    I can try to make it but only if x is an integer.

    Let's factorize :

    2^{\sqrt{x}} + 2^x = 2^{\sqrt{x}} ( 1+2^{x-\sqrt{x}}) = 20

    20 is 2≤*5
    1+2^{x-\sqrt{x}} is not an even number

    So 2^{\sqrt{x}} is 2≤

    -> \sqrt{x} = 2

    And you can conclude ^^
    Last edited by Moo; March 12th 2008 at 11:14 AM.
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  5. #5
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    Quote Originally Posted by perash View Post
    thank you mr fantastic
    but if
    2^{\sqrt{x}} = w
    then
    2^{2\sqrt{x}} = w^2
    not

    2^{{x}} = w^2
    Whoops!
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  6. #6
    o_O
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    Similar to what mr fantastic has done:

    2^{\sqrt{x}} + 2^{x} = 20
    2^{x^{\frac{1}{2}}} + 2^{x} = 20
    \left(2^{x}\right)^{\frac{1}{2}} + 2^{x} = 20

    Let w = 2^{x}

    w^{\frac{1}{2}} + w = 20
    w^{\frac{1}{2}} = 20 - w
    w = (20 - w)^{2}
    w = 400 - 40w + w^{2}
    0 = w^{2} - 41w + 400
    0 = (w - 16)(w - 25)
    w = 16 \quad \quad  w = 25
    2^{x} = 16 \quad \quad 2^{x} = 25

    Solve for x and ignore any erroneous answers.
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  7. #7
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    Quote Originally Posted by o_O View Post
    Similar to what mr fantastic has done:

    2^{\sqrt{x}} + 2^{x} = 20
    2^{x^{\frac{1}{2}}} + 2^{x} = 20
    \left(2^{x}\right)^{\frac{1}{2}} + 2^{x} = 20

    Mr F sadly adds: Unfortunately, (2^{x})^{1/2} = 2^{x/2}, not 2^{x^{1/2}} ......

    Let w = 2^{x}

    w^{\frac{1}{2}} + w = 20
    w^{\frac{1}{2}} = 20 - w
    w = (20 - w)^{2}
    w = 400 - 40w + w^{2}
    0 = w^{2} - 41w + 400
    0 = (w - 16)(w - 25)
    w = 16 \quad \quad  w = 25
    2^{x} = 16 \quad \quad 2^{x} = 25

    Solve for x and ignore any erroneous answers.
    Unfortunately that doesn't work either! But thanks for making me look less of a bonehead than I was!
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  8. #8
    Senior Member JaneBennet's Avatar
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    Let \mathrm{f}(x)=2^{\sqrt{x}}+2^x-20=\mathrm{e}^{\sqrt{x}\ln{2}}+\mathrm{e}^{x\ln{2}  }-20.

    Then \mathrm{f}'(x)=\mathrm{e}^{\sqrt{x}\ln{2}}\cdot\fr  ac{\ln{2}}{2\sqrt{x}}+\mathrm{e}^{x\ln{2}}\cdot\ln  {2}>0 for all x>0.

    Hence f is a strictly increasing function for all positive x; therefore it cannot cross the x-axis more than once. Since you have found a solution x=4, you can be sure there canít be any more.
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