Question

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Answer 1

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}