Question

Pessoal me ajuda ae a resolver as equações exponenciais:
a)9^x=$\sqrt[3]{3}$ b)8^x=$\sqrt[5]{2^4}$ c)1/9^x=$\sqrt{3}$

Answer 1

a) $9^{x}=\sqrt[3]{3}$

$\left(3^{2} \right )^{x}=3^{\,^{1}\!\!\!\diagup\!\!_{3}}\\ \\ 3^{2x}=3^{\,^{1}\!\!\!\diagup\!\!_{3}}\\ \\ 2x=\dfrac{1}{3}\\ \\ x=\dfrac{1}{3\cdot 2}\\ \\ x=\dfrac{1}{6}$


b) $8^{x}=\sqrt[5]{2^{4}}$

$\left(2^{3} \right )^{x}=2^{\,^{4}\!\!\!\diagup\!\!_{5}}\\ \\ 2^{3x}=2^{\,^{4}\!\!\!\diagup\!\!_{5}}\\ \\ 3x=\dfrac{4}{5}\\ \\ x=\dfrac{4}{5\cdot 3}\\ \\ x=\dfrac{4}{15}$


c) $\dfrac{1}{9^{x}}=\sqrt{3}$

$\dfrac{1}{\left(3^{2} \right )^{x}}=3^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ \dfrac{1}{3^{2x}}=3^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ \left(3^{2x} \right )^{-1}=3^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ 3^{2x\,\cdot \left(-1 \right )}=3^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ 3^{-2x}=3^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ -2x=\dfrac{1}{2}\\ \\ x=\dfrac{1}{2\cdot \left(-2 \right )}\\ \\ x=-\dfrac{1}{4}$