Question

Me ajudem a simplificar 8^(2-2log de raiz cúbica de 3 na base 4) + 1/3•7^log 4 na base 49)

Answer 1

Desenvolvendo:


$\large\begin{array}{l}8^{2-2\ell og_4\sqrt[3]{3}}+\dfrac{1}{3}\cdot 7^{\ell og_{49}4}~\Leftrightarrow~8^2\cdot8^{-2\ell og_4\sqrt[3]{3}}+\dfrac{1}{3} \cdot7^{\ell og_{7^2}4}\\\\8^2\cdot(2^3)^{-2\ell og_4\sqrt[3]{3}}}+\dfrac{1}{3}\cdot7^{\frac{1}{2}\ell og_74}~\Leftrightarrow~8^2\cdot2^{-6\ell og_4\sqrt[3]{3}}+\dfrac{7^{\ell og_74^{\frac{1}{2}}}}{3}\end{array}$

$\large\begin{array}{l}8^2\cdot2^{-6\ell og_{2^2}\sqrt[3]{3}}+\dfrac{7^{\ell og_7 \sqrt{4}}}{3}~\Leftrightarrow~8^2\cdot2^{-\frac{6}{2}\ell og_2\sqrt[3]{3} }+ \dfrac{7^{\ell og_72}}{3}\\\\\\8^2\cdot2^{-3\ell og_2\sqrt[3]{3}}+\dfrac{7^{\ell og_72}}{3}~\Leftrightarrow~8^2\cdot2^{-3\ell og_23^{ \frac{1}{3}}}+\dfrac{7^{\ell og_72}}{3}\end{array}$

$\large\begin{array}{l}8^2\cdot2^{\ell og_23^{-1}}}+\dfrac{7^{\ell og_72}}{3}~\Leftrightarrow~64\cdot 2^{\ell og_2(3^{-1})}+\dfrac{7^{\ell og_72}}{3} \end{array}$

Lembrando do seguinte caso:

$\Large\fbox{ a^{\ell og_a(b)}=b }$

O que temos é:

$\large\begin{array}{l}(64\cdot 3^{-1})+\dfrac{2}{3}~\Leftrightarrow~(64\cdot \dfrac{1}{3})+\dfrac{2}{3}~\Leftrightarrow~\dfrac{64}{3}+\dfrac{2}{3}~\Leftrightarrow~\dfrac{66}{3}=\fbox{ 22 }\end{array}$

Portanto:

$\large\begin{array}{l}\fbox{ 8^{2-2\ell og_4\sqrt[3]{3}}+\dfrac{1}{3}\cdot 7^{\ell og_{49}4}~\Longleftrightarrow~22 }\end{array}$