1. ## 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) $\displaystyle (1-a)^n<a^n\ \ (n\in\mathbb{N})$

2. Originally Posted by math sucks
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.