The expression is [tex]3\log_bm-2\log_bn[/tex].
Recall the following rules:
i) [tex]\displaystyle{ \log_ba-\log_bc=\log_b{\frac{a}{c}}[/tex]
ii) [tex]\displaystyle{ a\log_bc= \log_bc^a[/tex]
First, by the rule (ii) we have:
[tex]3\log_bm-2\log_bn=\log_bm^3-\log_bn^2.[/tex]
Second, by rule (i), we have:
[tex]\displaystyle{ \log_bm^3-\log_bn^2=\log_b{\frac{m^3}{n^2}}.[/tex]
Thus, the answer is : [tex]\displaystyle{ \log_b{\frac{m^3}{n^2}}[/tex]
