Question

será que alguém pode me AJUDAR

$log_{4} \sqrt{32}$
POR FAVOR ME AJUDEM ​

Answer 1

Resposta:

$\log_4\sqrt{32} =x\\ \\ 4^x=\sqrt{2^5} \\ \\ (2^2)^x=2^{{5\over2}}\\ \\ 2^{2x}=2^{{5\over2}}\\ \\ cancela~~2\\ \\ 2x={5\over2}\\ \\ 4x=5\\ \\ x={5\over4}\\ \\ logo\\ \\ \log_4\sqrt{32} ={5\over4}$

Answer 2

$log_{4}(\sqrt{32})=log_{{2}^{2}}({2}^{\frac{5}{2}}) \\=\dfrac{\dfrac{5}{2}}{2}=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}$