Question

Log4 √32 nunca consigo fazer quando tem raiz, me ajudem

Answer 1

Log 4 $\sqrt{32}$=x

4^x= $\sqrt{32}$

2^2x=$\sqrt{2^5}$


2^2x=2^5/2

2x=5/2

x=5/2 / 2

x=5/2*1/2

x=5/4

Answer 2

$\large\begin{array}{l} \textsf{Para calcular logaritmos envolvendo raiz, voc\^e pode}\\\textsf{transformar para expoente fracion\'ario. Observe:}\\\\ \mathsf{\ell og_4\,\sqrt{32}=x}\\\\ \mathsf{4^x=\sqrt{32}}\\\\ \mathsf{4^x=(32)^{\frac{1}{2}}}\\\\ \mathsf{(2^2)^x=(2^5)^{\frac{1}{2}}} \end{array}$

$\large\begin{array}{l} \mathsf{(2^2)^x=(2^5)^{\frac{1}{2}}}\\\\ \mathsf{2^{2x}=2^{5\,\cdot\,\frac{1}{2}}}\\\\ \mathsf{2^{2x}=2^{\frac{5}{2}}} \end{array}$


$\large\begin{array}{l} \textsf{Agora, temos uma igualdade entre exponenciais de mesma base.}\\\textsf{Basta igualar os expoentes:}\\\\ \mathsf{2x=\dfrac{5}{2}}\\\\ \mathsf{x=\dfrac{5}{2}\cdot \dfrac{1}{2}}\\\\ \mathsf{x=\dfrac{5}{4}}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\ell og_4\,\sqrt{32}=\dfrac{5}{4}} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$