(a) 5+(−3)−21=23.
(b) loga2a3=23 e log1/aa4=−4, quindi 23−4=−25.
(c) loga(a3b2)=3+2logab=3+6=9; logbb4=4; logcc3=3: 39⋅4=12.
(d) (a2)loga2=(aloga2)2=22=4; clogc3+1=clogc3⋅c=3c. Risultato 4−3c: dipende da c (numerico solo se il termine è clogc3=3, nel qual caso vale 1).
(e) logac=31, logab=21: loga(c2b3)=32+23=613; logb(c/a)=logbc−2=32−2=−34. Totale 613+34=621=27.
(a) 23; (b) −25; (c) 12; (d) 4−3c; (e) 27