Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):

$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$
$$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$

Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?