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    Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!
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    (Original post by noodlemon)
    Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!
    What have you tried?
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    (Original post by noodlemon)
    Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!
    Hint:

    Let

    f(x)= x-\ln(x+5)

    and notice that if, at x=a, f(x) crosses the x axis, it must be above on one side of a and below on the other.
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    (Original post by TenOfThem)
    What have you tried?
    ok ive tried getting rid of the ln getting x+5=e^x? i suppose you need to make x the subject now right?
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    (Original post by noodlemon)
    ok ive tried getting rid of the ln getting x+5=e^x? i suppose you need to make x the subject now right?
    Nope

    Did you try putting 1 and 2 in to see what happens
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    (Original post by Indeterminate)
    Hint:

    Let

    f(x)= x-\ln(x+5)

    and notice that if, at x=a, f(x) crosses the x axis, it must be above on one side of a and below on the other.
    so are you saying to use trial and error?
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    (Original post by noodlemon)
    so are you saying to use trial and error?
    You are not trying to find x

    RTQ
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    (Original post by TenOfThem)
    You are not trying to find x

    RTQ
    Oh i see now, sorry ah good advice.. i subbed in 1 to the equation and got 0.7917... and 2 to get -0.054.. meaning it must have crossed the x axis between those points of 1 and 2. thanks
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    (Original post by noodlemon)
    so are you saying to use trial and error?
    Well, no.

    \displaystyle x=\ln(x+5) \Rightarrow x-\ln(x+5)=0

    Thus if you let f(x) equal the LHS, you need to show that there is a root of f(x)=0 between 1 and 2 (i.e it crosses the x axis)

    Use the fact that, to cross the x-axis, it must be above the x axis at one point (on one side of the root), and below on the other.

    See if you can relate what I'm saying to the values of f(x) when you substitute 1 and 2 in
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    (Original post by Indeterminate)
    Well, no.

    \displaystyle x=\ln(x+5) \Rightarrow x-\ln(x+5)=0

    Thus if you let f(x) equal the LHS, you need to show that there is a root of f(x)=0 between 1 and 2 (i.e it crosses the x axis)

    Use the fact that, to cross the x-axis, it must be above the x axis at one point (on one side of the root), and below on the other.

    See if you can relate what I'm saying to the values of f(x) when you substitute 1 and 2 in
    yup aha thanks guys i was overcomplicating the question and trying to find the exact value of x :3
 
 
 
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