Return to Question

For any cardinal $\kappa$ set $$\log(\kappa) = \min\{\mu\in \kappa: 2^\mu \geq \kappa\}.$$ 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 $\kappa$?

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 $\kappa$?

For any cardinal $\kappa$ set $$\log(\kappa) = \min\{\mu: 2^\mu \text{ is a cardinal }\land \geq \kappa\}.$$ 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 $\kappa$?

For any cardinal $\kappa$ set $$\log(\kappa) = \min\{\mu\in \kappa: 2^\mu \geq \kappa\}.$$ 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 $\kappa$?