Suppose that u is subharmonic in the plane and that, for some p > 1 , lim _ r → ∞ B ( r ) / ( log ⁡ r ) p = σ > ∞ p > 1,{\underline {\lim } _{r \to \infty }}B(r)/{(\log r)^p} = \sigma > \infty . It is shown that, given ε > 0 \varepsilon > 0 , \[ A ( r ) > B ( r ) − ( σ + ε ) Re ⁡ { ( log ⁡ r ) p − ( log ⁡ r + i π ) p } A(r) > B(r) - (\sigma + \varepsilon )\operatorname {Re} \{ {(\log r)^p} - {(\log r + i\pi )^p}\} \] for r outside an exceptional set E, where \[ lim _ x → ∞ 1 ( log ⁡ r ) p − 1 ∫ E ∩ [ 1 , r ] ( p − 1 ) ( log ⁡ t ) p − 2 t d t ⩽ σ σ + ε . \underline {\lim } \limits _{x \to \infty } \;\frac {1}{{{{(\log r)}^{p - 1}}}}\int _{E \cap [1,r]} {\frac {{(p - 1){{(\log t)}^{p - 2}}}}{t}\;dt\; \leqslant \frac {\sigma }{{\sigma + \varepsilon }}.} \]