How to prove that $5^q<7^q$ implies $q>0$? Consider $q\in \mathbb{Q}\:$ , $5^{q\:}<\:7^q$
How to prove that $q>0$ using only Power rules? 
So far, I just know that if $q=0$ we get $1>1$ and its false. But what to do get contradict for $q< 0$?
 A: First, note that for $q > 0$ we have
$$a > 1 \Rightarrow a^q > 1 \text{ and } 0 < a < 1 \Rightarrow 0 < a^q < 1.$$
Hence, for $q > 0$ it follows that
$$7^q = \left( 5 \cdot \frac{7}{5} \right)^q = 5^q \cdot \underbrace{\left( \frac{7}{5} \right)^q}_{>1} > 5^q \cdot 1 = 5^q,$$
i.e. the inequality is true for $q > 0$. As you already noted, it is false for $q = 0$. Now assume $q < 0$. Then $q = - q'$ where $q' > 0$ and we have
$$ 7^q = 7^{-q'} = \frac{1}{7^{q'}} = \frac{1}{\left( 5 \cdot \frac{7}{5} \right)^{q'}} = \frac{1}{5^{q'}} \cdot \underbrace{\left( \frac{5}{7} \right)^{q'}}_{< 1} < \frac{1}{5^{q'}} = 5^{-q'} = 5^q.$$ 
It follows that $q < 0$ implies $7^q < 5^q$.
Now if you assume that $7^q > 5^q$, then it follows that $q > 0$, because the inequality is false for $q = 0$ and $q<0$.
Sorry if this is a bit long.
A: $5^q<7^q \implies 1<\left(\dfrac{7}{5}\right)^q\implies\left(\dfrac{7}{5}\right)^0<\left(\dfrac{7}{5}\right)^q \implies q>0$
A: $$5^q<7^q$$
Since the function $\log_5(\cdot)$ is strictly increasing,
$$q<q\log_57$$
or
$$q(\log_57-1)>0$$
and since $\log_57>\log_55=1$,
$$q>0$$
