Is there a gap in my proof? If $a>1$ and $\log_ab_1>\log_ab_2$, then $b_1>b_2.$ I need to prove that

If $a>1$ and $\log_ab_1>\log_ab_2$, then $b_1>b_2.$

My attempts:

Let $\log_ab_1=x, \log_ab_2=y$ we have $\begin{cases} a^x=b_1 \\ a^y=b_2 \end{cases} \Longrightarrow a^{x-y}=\dfrac{b_1}{b_2}$
$a^{x-y}>a^0=1$, because $a>1$, then $\dfrac{b_1}{b_2}>1 \Longrightarrow  b_1>b_2.$

This argument works for $a<1$ and $\log_ab_1<\log_ab_2$ too.
Is there a gap in my attempts?
Thank you.
 A: To address your "gap in proof" question, I agree with Emilio that the only original gap was not stating that $a^{x-y}>a^0=1$ because $a>1$ (but you have now edited this observation into your question post). Since the exponential function increases only if the base is greater than 1, your argument is fine as is. Good work.
I think the argument could be made clearer/cleaner by working out something like what I have included below. You're not introducing any new variables or making any real jumps between observations. 

I would first note that $\log x > 0$ when $x>1$ and $10^x$ is a strictly increasing function. Hence,
\begin{align}
\log_a b_1>\log_a b_2
&\implies \frac{\log b_1}{\log a}>\frac{\log b_2}{\log a} & \text{(change of base)}\\[1em]
&\implies \log b_1 > \log b_2 & \text{(since $a>1$)}\\[1em]
&\implies 10^{\log b_1}>10^{\log b_2} & \text{(since $10^x$ strictly increases)}\\[1em]
&\implies b_1 > b_2. &\text{(inverse functions)}
\end{align}
Hopefully that helps guide your thinking a bit.
A: Hint:
$x>y \quad \Rightarrow a^{x-y}>1$
only if $a>1$.
