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  1. #1
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    solve

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

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

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

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

    Therefore:

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

    $\displaystyle 2^{\sqrt{x}} = -5$ 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 $\displaystyle 2^{\sqrt{x}} = w$. Then:

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

    Therefore:

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

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

    $\displaystyle 2^{{x}} = w^2$
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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 :

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

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

    So $\displaystyle 2^{\sqrt{x}}$ is 2≤

    -> $\displaystyle \sqrt{x} = 2$

    And you can conclude ^^
    Last edited by Moo; Mar 12th 2008 at 10:14 AM.
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  5. #5
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    Quote Originally Posted by perash View Post
    thank you mr fantastic
    but if
    $\displaystyle 2^{\sqrt{x}} = w$
    then
    $\displaystyle 2^{2\sqrt{x}} = w^2$
    not

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

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

    Let $\displaystyle w = 2^{x}$

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

    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:

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

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

    Let $\displaystyle w = 2^{x}$

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

    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
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    Let $\displaystyle \mathrm{f}(x)=2^{\sqrt{x}}+2^x-20=\mathrm{e}^{\sqrt{x}\ln{2}}+\mathrm{e}^{x\ln{2} }-20$.

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

    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 $\displaystyle x=4$, you can be sure there canít be any more.
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