Question

㏒$_{8}$ 2√2

Answer 1

$\mathsf{\ell og_{8}\,(2\sqrt{2})=x~~\Rightarrow~~8^{x}=2\sqrt{2}}\\\\ \mathsf{(2^{3})^{x}=2\cdot 2^{1/2}}\\\\ \mathsf{2^{3x}=2^{1}\cdot 2^{1/2}}\\\\ \mathsf{2^{3x}=2^{1+\frac{1}{2}}}\\\\ \mathsf{2^{3x}=2^{\frac{3}{2}}}$


Na última linha acima, temos uma igualdade entre exponenciais de mesma base. Então é só igualar os expoentes:

$\mathsf{3x=\dfrac{3}{2}}\\\\\\ \mathsf{x=\dfrac{(\frac{3}{2})}{3}}\\\\\\ \mathsf{x=\dfrac{\diagup\!\!\!\! 3}{2}\cdot\dfrac{1}{\diagup\!\!\!\! 3}}\\\\\\ \mathsf{x=\dfrac{1}{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\ell og_{8}\,(2\sqrt{2})=\dfrac{1}{2}} \end{array}}$