Question

URGENT
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Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Vom folosi două formule generale:

$\it a^{\frac{1}{2}} =\sqrt a\\ \\ k\cdot log_bx=log_bx^k$

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$\it \log_2 3-\dfrac{1}{2}log_2 9 = log_2 (3^2)^{\frac{1}{2}}-\dfrac{1}{2}log_2 9=\dfrac{1}{2}log_29-\dfrac{1}{2}log_2 9 = 0$