Calculate maximum for $(\log_{7}4)^{\sin x}$ How do I calculate maximum of: :$$(\log_{7}4)^{\sin x}$$
NOTE:
1) I know Basical rules for logarithms
2)  I want you to hint me (somehow) completely through the path to solve the problem! Thanks.
 A: As $0<\log_74<1,$ the given expression will attain maximum if  $\sin x$ is minimum
A: We know that $0<\log_4{7}<1$. Therefore the only way to maximize it is to minimize the power, $\sin(x)$.
$\sin(x)$ is minimized when it equals -1. This occurs at $x=\frac{3\pi}{2}+2\pi n$, $n\in\Bbb Z$, which is your answer.
A: Hint:
For any $a$, $a^x$ is a strictly monotonous function, hence it preserves the extrema.
A: Key points:
1) If $0 < a; a\ne 1$ then $a^x$ is either monotonically increasing (if $a > 1$) or monotonically decreasing (if $a < 1$).

So $\max a^{f(x)} = a^{\max f(x)}$ if $a > 1$ or $\max a^{f(x)} = a^{\min f(x)}$ if $a < 1$.

2) If $1 < a < b$ then $\log_a b = k$ means $a^k = b > a$ so $k =\log_a b > 1$.  If $1 < b < a$ then $\log_a b= k$ means $1 < a^k = b < a $ and $0 < k = \log_a b < 1$.

So If $1 < a < b$ then $\log_a b > 1$.  If $1 < b < a$ then $0 < \log_a b < 1$.

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So $7 > 4$ so $0 < \log_7 4 < 1$.
So $\max (\log_7 4)^{\sin x}= (\log_7 4)^{\min \sin x}$.
And $\min \sin x = -1$
So $\max (\log_7 4)^{\sin x} =(\log_7 4)^{-1} = \frac 1{\log_7 4}$.
And if $a^k = b$ then $a= b^{\frac 1k}$ 

so $\log_b a = \frac 1{\log_a b}$.

So $\max (\log_7 4)^{\sin x} = \log_4 7$.
