The function lnx

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    Is this right?

    e^2x+1 = 2
    2x +1 = ln 2
    x = 0.5 ln( 2- 1)

    I though for the last step:

    x = 0.5 (ln(2) -1)

    with the one outside the ln brakets?
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    You are correct. The last step is x = 0.5 (ln(2)-1)
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    (Original post by CharlieBoardman)
    You are correct. The last step is x = 0.5 (ln(2)-1)
    Thats what is thought but the mark sheme says this

    0.5 ln( 2- 1)
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    (Original post by nazgul60)
    Thats what is thought but the mark sheme says this

    0.5 ln( 2- 1)
    That's wrong.
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    (Original post by nazgul60)
    Thats what is thought but the mark sheme says this

    0.5 ln( 2- 1)
    It is wrong because that gives 0.5ln(1) = 0 and this can be easily disproved by substituting x=0 into the initial equation.
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    (Original post by Micky76)
    It is wrong because that gives 0.5ln(1) = 0 and this can be easily disproved by substituting x=0 into the initial equation.
    Thats what i thought but just needed to be confident. Thanks!
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    Am I just being stupid or isn't x=0 the answer?
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    e^{2x+1}=2

    or

    e^{2x}+1=2
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    (Original post by BabyMaths)
    e^{2x+1}=2

    or

    e^{2x}+1=2
    Is it not multiplied by e instead of the +1?
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    (Original post by BabyMaths)
    e^{2x+1}=2


    that one
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    (Original post by BabyMaths)
    e^{2x+1}=2

    or

    e^{2x}+1=2
    Sorry, my bad in the interpretation of your post.
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    (Original post by nazgul60)
    that one
    In that case x=\frac{1}{2}(-1+\ln 2) as you thought.
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    (Original post by nazgul60)
    Thats what is thought but the mark sheme says this

    0.5 ln( 2- 1)
    Sometimes mark schemes have mistakes in them. Don't worry - you are correct!
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    (Original post by nazgul60)
    that one
    Then you are right/mark scheme is wrong.

    probably just a typing error by a clerk who doesn't know to be careful with these things.
 
 
 
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