Explanation

if $a = log^5_3$ and $b = log_{27}^{25}$ = $log^{5^2}_{3^3}$ = $\frac {2}{3} log^5_3$

but $a = log^5_3$ ::: $b = \frac {2}{3}a$

hence $a>b$

if $a = log^5_3$ and $b = log_{27}^{25}$ = $log^{5^2}_{3^3}$ = $\frac {2}{3} log^5_3$

but $a = log^5_3$ ::: $b = \frac {2}{3}a$

hence $a>b$