Question

Determine o valor de x na equação exponencial 9 elevado a x+3=(1/27)³

Answer 1

$9^{x+3}=\left(\dfrac{1}{27}\right)^{3}\\\\\\(3^{2})^{x+3}=\left(\dfrac{1}{3^{3}}\right)^{3}\\\\\\3^{2\cdot(x+3)}=\dfrac{1}{3^{3\cdot3}}\\\\\\3^{2x+6}=\dfrac{1}{3^{9}}$

Invertendo a base e trocando o sinal do expoente:

$3^{2x+6}=3^{-9}$

Bases iguais, iguale os expoentes:

$2x+6=-9\\2x=-9-6\\2x=-15\\\\\boxed{\boxed{x=-\dfrac{15}{2}}}$