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Thread: Confusing Logarithms

  1. #1
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    Confusing Logarithms

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

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

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

    (A)

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

    True.

    (B)

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

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

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

    False.

    (C)

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

    (D)

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

    False.
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