To prove: log_a(uv) = log_au + log_av

a^xa^y = a^x+y implies

log_a(a^xa^y) = log_a(a^x+y)

log_a(a^xa^y) = x+y

Set x = log_au, y = log_av

log_a(a^logaua^logav) = log_au + log_av

log_a(uv) = log_au + log_av