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For any cardinal $\kappa$ set $$\log(\kappa) = \min\{\mu\in \kappa\cup\{\kappa\}: 2^\mu \geq \kappa\}.$$ Clearly $\log(\omega) = \omega$ and in $\textsf{GCH}$ we have $\log(\aleph_\omega) = \aleph_\omega$.

Is it consistent that $\log(\kappa)<\kappa$ for all uncountable cardinals $\kappa$?

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No. Because strong limits cardinals exist in $\sf ZFC$. These are the $\beth_\alpha$ for a non-successor $\alpha$.

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  • $\begingroup$ And without choice? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 17, 2017 at 10:14
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    $\begingroup$ You have to be more careful, but there is a reasonable sense in which the same statement is provable without choice. $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 17, 2017 at 10:59

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