Question

AJUDEM PORFAVOR URGENTE:

Ache y real sabendo-se que: $log_{2}$ y = $log_{2} 3$ + $log_{2} 6$ - 3$log_{2} 4$

Answer 1

$\mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2\,}3+\mathrm{\ell og}_{2\,}6-3\mathrm{\,\ell og}_{2\,}4\\ \\ \mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2\,}3+\mathrm{\ell og}_{2\,}6-\mathrm{\ell og}_{2}\left(4^{3}\right)\\ \\ \mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2\,}3+\mathrm{\ell og}_{2\,}6-\mathrm{\ell og}_{2\,}64\\ \\ \mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2}\left(\dfrac{3 \cdot 6}{64} \right )\\ \\ \mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2\,}\dfrac{18}{64}\\ \\ \mathrm{\ell og}_{2\,}y=\mathrm{\ell og}_{2\,}\dfrac{9}{32}\\ \\ \boxed{y=\dfrac{9}{32}}$