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    A root of the equation e^x-7x=0 lies in the interval N<x<N+1 where N is an integer. Find 2 values for N.

    I thought maybe iteration would do the trick, but I certainly didn't get the answers in the book with it...
    So I figured hey maybe I can log both sides, but well you get ln 7x which is weird as after ln 7 + ln x I am stuck...
    Any ideas?
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    (Original post by BossLady)
    A root of the equation e^x-7x=0 lies in the interval N<x<N+1 where N is an integer. Find 2 values for N.

    I thought maybe iteration would do the trick, but I certainly didn't get the answers in the book with it...
    So I figured hey maybe I can log both sides, but well you get ln 7x which is weird as after ln 7 + ln x I am stuck...
    Any ideas?
    first consider the graphs of y=e^x and y=7x and from there x must be >0,then let f(x)=e^x-7x and start with x=0 until u get a sign change....
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    (Original post by IntegralAnomaly)
    first consider the graphs of y=e^x and y=7x and from there x must be >0,then let f(x)=e^x-7x and start with x=0 until u get a sign change....
    you mean put x=0 in??
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    (Original post by BossLady)
    you mean put x=0 in??
    yea, cos f(x)=e^x-7x
    f(0)=1-7=-6
    then since its asking for an integer take (f1)
    f(1)=-11.3
    still negative so lets try x=2
    f(2)=-6.61
    still negative so.........
    f(3)=-0.91 hmmm
    f(4)=26.59
    wolla! a sign change so 1 value for N is N=3. Now get the other 1.....
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    woops my mistake f(0)=1 and f(1)=-4.28.
    a sign change so N=0 and N=3
 
 
 

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