pigor: ..., no to jeszcze dowód: a>0, c>0 i b>0 i b≠1
i niech a
log bc= x i c
log ba= y i x>0 i y>0 (na pewno),
to logarytmując obustronnie logarytmem o podstawie b ⇔
⇔ log
b(a
log bc) = log
bx i log
b(c
log ba) = log
by ⇔
⇔
log bc * log ba = log
bx i
log ba * log bc = log
by ⇒
⇒ log
bx = log
by ⇔ x = y ⇔
alog bc = clog ba c.n.u.