How to prove that $5^q<7^q$ implies $q>0$?

Question

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 $q< 0$?

Answer 1

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.

Answer 2

$$5^q<7^q$$ Since the function $\log_5(\cdot)$ is strictly increasing, $$q<q\log_57$$ or $$q(\log_57-1)>0$$ and since $\log_57>\log_55=1$, $$q>0$$

Answer 3

$5^q<7^q \implies 1<\left(\dfrac{7}{5}\right)^q\implies\left(\dfrac{7}{5}\right)^0<\left(\dfrac{7}{5}\right)^q \implies q>0$