$|b|^{\frac {1}{\log b}}= |a|^{\frac {1}{\log a}}$ implies $|b|=b^ {\lambda}$

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I was studying 'Ostrowski's Theorem' stating that "Every non-trivial valuation on $\Bbb Q$ is equivalent either to a $p-$adic valuation or to the ordinary absolute value.

While reading the proof I have reached a point where the book says that for all integers $a,b >1$ if we have $$|b|^{\frac {1}{\log b}}= |a|^{\frac {1}{\log a}}$$

where $||$ is a valuation on $\Bbb Q$.

Then it follows that $$|b|=b^ {\lambda}$$ for all integer $b>1$ and for some $\lambda$ independent of $b$.

I can't solve this line of reason. Please help.

Book is "Local Fields" by J.W.S Cassels

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Let $c:=|b|^{\frac {1}{log b}}= |a|^{\frac {1}{log a}}.$ Then we have $$ |b|=c^{\log(b)}=e^{\log(c)\log(b)}=b^{\log(c)}=b^{\lambda}. $$ Hence $|b|=|b|_{\infty}^{\lambda}$, where $|b|_{\infty}$ is the usual absolute value on $\mathbb{Q}$.