Ex 1
1) log3 (243) = log3 (3⁵) = 5 log3 (3) = 5 *1 = 5
2) 1 - 0 + log2 (16) = 1+ log2 (2⁴) = 1+4 = 5
3) log7 (42/6) = log7 (7) = 1
4) log5 (6*12) = log5 (72)
5) log6 (18*4)/12 = log6 (18/3) = log6 (6) = 1
6) log2 ((V5-V3)(V5+V3)) = log2 (V5² - V3²) = log2 (5-3) = log2 (2) = 1
Ex 2 Aplici conditiile de existenta la logaritmului
a) CE x+5 > 0
x > -5
x apartine (-5, infinit)
b) CE x-2 # 1 , x # 3
x-2 > 0, x > 2
x apartine (2, infinit) /{3}
