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Math Help - Special equation?

  1. #1
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    Special equation?

    Wrong forum probably, but I'm not sure where to post...
    My friend challenged me with this curious equation:

    2^x - 5 = x

    Like I know the answer is 3, but I don't know how to mathematically prove this.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    Wrong forum probably, but I'm not sure where to post...
    My friend challenged me with this curious equation:

    2^x - 5 = x

    Like I know the answer is 3, but I don't know how to mathematically prove this.
    All else fails I always do this

    f(x)\equiv{2^x-5-x}

    so f(2)=-3

    and f(4)=7

    Therefore by the intermediate value theorem there \exists{c}\in(2,4)\backepsilon{f(c)}=0

    So now utilizing probably three but for the sake of the absence of luck in math

    say the starting point for Newton-Raphson method is 2.5

    a_0=2.5

    a_2=2.5-\frac{f(2.5)}{f'(2.5)}

    a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}

    we see that as n\to\infty a_n\to{3}


    The check step

    f(3)=0


    Sorry that is bad but I cannot think of anything else as of present


    it works though
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  3. #3
    MHF Contributor arbolis's Avatar
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    My friend challenged me with this curious equation:

    2^x - 5 = x
    It might looks simple, but it's a real challenge. In fact, if f(x)=2^x-5-x, then f has 2 zeros. It means that the solution is not singular. But to prove it... not easy.
    f'(x)=ln(2)e^{xln(2)}=ln(2)e^x-1. We want to study its sign. So we want to find when it is equal to 0. f'(x)=0 \Leftrightarrow ln(2)e^x=1 \Leftrightarrow x=ln(\frac{1}{ln(2)})=-ln(ln2).
    Now if x<-ln(ln2), f'(x)<0. It means that f(x) is decreasing from negative infinite to -ln(ln2). As when x>-ln(ln2), f'(x)>0, f(x) is increasing from -ln(ln2) to positive infinite.
    f(0)=-4. As it will increase (strictly), it must cross the x axis on a single point when x>0. You can find a similar conclusion calculating for example f(-10). It means that the roots of f are respectively between [-10,0] and [0,5]. (5 because f(5)>0). Now to find the roots, you can say that there is an easy one to see, when x=3. To calculate the other root, I suggest you the bisection method which will work.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    It might looks simple, but it's a real challenge. In fact, if f(x)=2^x-5-x, then f has 2 zeros. It means that the solution is not singular. But to prove it... not easy.
    f'(x)=ln(2)e^{xln(2)}=ln(2)e^x-1. We want to study its sign. So we want to find when it is equal to 0. f'(x)=0 \Leftrightarrow ln(2)e^x=1 \Leftrightarrow x=ln(\frac{1}{ln(2)})=-ln(ln2).
    Now if x<-ln(ln2), f'(x)<0. It means that f(x) is decreasing from negative infinite to -ln(ln2). As when x>-ln(ln2), f'(x)>0, f(x) is increasing from -ln(ln2) to positive infinite.
    f(0)=-4. As it will increase (strictly), it must cross the x axis on a single point when x>0. You can find a similar conclusion calculating for example f(-10). It means that the roots of f are respectively between [-10,0] and [0,5]. (5 because f(5)>0). Now to find the roots, you can say that there is an easy one to see, when x=3. To calculate the other root, I suggest you the bisection method which will work.
    Haha stupid two roots
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