Question

3.4(x+1)=96 como resolve? Funções exponenciais :)

Answer 1

$3\cdot4^{x+1}=96\\\\ (2^2)^{x+1}= \dfrac{96}{3}\\\\ 2^{2x+2}=32\\ 2^{2x+2}=2^5\\ \not2^{2x+2}=\not2^5\\\\ 2x+2=5\\ 2x=3\\\\ x= \dfrac{3}{2}\\\\\\ \Large\boxed{\text{S}=\left\{ \dfrac{3}{2}\right\}}$