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Math Help - Confusing Logarithms

  1. #1
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    Confusing Logarithms

    Let 0<a<\dfrac{1}{2}, then which of the following must be true

    (A) \log_a(1-a)>1
    (B) \log_a(1-a)<\log_{1-a}a
    (C) a^{1-a}>(1-a)^a
    (D) (1-a)^n<a^n\ \ (n\in\mathbb{N})
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  2. #2
    Member SengNee's Avatar
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    Pangkor Island, Perak, Malaysia.
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    Quote Originally Posted by math sucks View Post
    Let 0<a<\dfrac{1}{2}, then which of the following must be true

    (A) \log_a(1-a)>1
    (B) \log_a(1-a)<\log_{1-a}a
    (C) a^{1-a}>(1-a)^a
    (D) (1-a)^n<a^n\ \ (n\in\mathbb{N})
    I just try...

    (A)

    \log_a(1-a)>1
    1-a>a
    a<\frac{1}{2}

    True.

    (B)

    \log_a(1-a)<\log_{1-a}a
    \log_a(1-a)<\frac{1}{log_{a}{1-a}}
    \left[\log_a(1-a)\right]^{2}<1
    \left[\log_a(1-a)\right]^{2}-1<0
    \left[\log_a(1-a)+1\right]\left[\log_a(1-a)-1\right]<0

    \log_a(1-a)<-1
    1-a<a^{-1}
    a-a^{2}<1
    0<a^{2}-a+1
    b^2-4ac<0
    Therefore no real roots.

    \log_a(1-a)<1
    1-a<a
    a>\frac{1}{2}

    False.

    (C)

    Confused..............

    (D)

    (1-a)^n<a^n
    n\log_{a}{1-a}<n\log_{a}{a}
    \log_{a}{1-a}<1
    1-a<a
    a>\frac{1}{2}

    False.
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