Question

resolva a equaçao

$\sqrt[4]{9 ^x^-^2} = \sqrt[3]{27^x^+^3}$

Answer 1

$\\\sqrt[4]{9^{x-2}}=\sqrt[3]{27^{x+3}}\\\\9^{\frac{x-2}{4}}=27^\frac{x+3}{3}\\\\\left(3^2 \right)^{\frac{x-2}{4}}=\left(3^3\right)^\frac{x+3}{3}\\\\\left(3\right)^\frac{2(x-2)}{4}=\left(3\right)^\frac{3(x+3)}{3}\\\\\frac{(x-2)}{2}=\frac{(x+3)}{1}\\\\x-2=2x+6\\\\x-2x=6+2\\\\\boxed{x=-8}$

Answer 2

$\sqrt[4]{9 ^{x-2} } = \sqrt[3]{27 ^{x+3} }$

$9 x^{\frac{x-2}{4}}=27 ^{\frac{x+3}{3} }$

$3 ^{\frac{2x-4}{4} } =3^{\frac{3x+9}{3} }$

$\frac{2x-4}{4}= \frac{3x+9}{3}$

$\frac{2(x-2)}{4}=\frac{3(x+3)}{3}$

$\frac{x-2}{2} =x+3$

$x-2=2x+6$

$-x=8$

$x=-8$