Question

como resolver essa equação exponencial :
9x+1 = $\sqrt[3]{3}$

Answer 1

Propriedades usadas:

$(a^{x})^{y}=a^{(x\cdot y)}\\\sqrt[y]{a^{x}}=a^{x/y}$
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$9^{x+1}=\sqrt[3]{3}\\(3^{2})^{x+1}=\sqrt[3]{3^{1}}\\3^{2(x+1)}=3^{1/3}$

Bases iguais, iguale os expoentes:

$2(x+1)=\frac{1}{3}\\\\x+1=\frac{1}{2\cdot3}\\\\x+1=\frac{1}{6}\\\\x=\frac{1}{6}-1\\\\x=\frac{1-6}{6}\\\\\\\boxed{\boxed{x=-\dfrac{5}{6}}}$