Question

Resolver equação exponencial:

Answer 1

$\sqrt[5]{2^x}\cdot\sqrt[3]{4^x}=\sqrt{8^{-x}}~\Longleftrightarrow~2^{\frac{x}{5}}\cdot 4^{\frac{x}{3}}=8^{-\frac{x}{2}}~\Longleftrightarrow~2^{\frac{x}{5}}\cdot 2^{\frac{2x}{3}}=2^{-\frac{3x}{2}}\\\\\Longleftrightarrow~2^{\frac{13x}{15}}=2^{-\frac{3x}{2}}~\Longleftrightarrow~\dfrac{13x}{15}=-\dfrac{3x}{2}~\Leftrightarrow~26x=-45x~\Longleftrightarrow~71x=0\\\\\Longleftrightarrow~x=0$

Answer 2

Explicação passo-a-passo:

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{4^x}=\sqrt{8^{-x}}$

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{(2^2)^x}=\sqrt{(2^3)^{-x}}$

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{2^{2x}}=\sqrt{2^{-3x}}$

$\sf 2^{\frac{x}{5}}\cdot2^{\frac{2x}{3}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{x}{5}+\frac{2x}{3}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{3x+10x}{15}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{13x}{15}}=2^{\frac{-3x}{2}}$

Igualando os expoentes:

$\sf \dfrac{13x}{15}=\dfrac{-3x}{2}$

$\sf \dfrac{13x}{15}+\dfrac{3x}{2}=0$

$\sf 26x+45x=0$

$\sf 71x=0$

$\sf x=\dfrac{0}{71}$

$\sf \red{x=0}$

O conjunto solução é $\sf S=\{0\}$