Question

resolver o sistema: 2^x*8^y=32
log(xy)na base 8=1/3

Answer 1

$\begin{Bmatrix}2^x*8^y=32\\\\\log_8(xy)=\frac{1}{3}\end{matrix}$

$\begin{Bmatrix}2^x*(2^3)^y=2^5\\\\xy=8^{\frac{1}{3}}\end{matrix}$

$\begin{Bmatrix}2^{x+3y}=2^5\\\\xy=2\end{matrix}$

$\begin{Bmatrix}x+3y=5\\\\y=\frac{2}{x}\end{matrix}$

$\begin{Bmatrix}x+3*\frac{2}{x}=5\\\\y=\frac{2}{x}\end{matrix}$

$\begin{Bmatrix}x^2+6=5x\\\\y=\frac{2}{x}\end{matrix}$

$\begin{Bmatrix}x^2-5x+6=0\\\\y=\frac{2}{x}\end{matrix}$

$\begin{Bmatrix}x_1=2~~e~~x_2=3\\\\y=\frac{2}{x}\end{matrix}$

$\boxed{\boxed{\begin{Bmatrix}x_1=2~~e~~x_2=3\\\\y_1=\frac{2}{2}=1~~e~~y_2=\frac{2}{3}\end{matrix}}}$

Answer 2

$2^x*8^y=32\Rightarrow 2^x*2^{3y}=2^5\Rightarrow 2^{x+3y}=2^5\Rightarrow x+3y=5\Rightarrow x=5-3y\\ \\ log_8(xy)=\frac{1}{3}\Rightarrow xy=8^{\frac{1}{3}} \Rightarrow xy= \sqrt[3]{8}\Rightarrow xy=2\\ \\ (5-3y)y=2\\ \\ 5y-3y^2-2=0\\ \\ -3y^2+5y-2=0\\ \\ y=\frac{2}{3} \ \ ou \ \ y=1\\ \\ Para \ y=\frac{2}{3} \Rightarrow x = 3\\ \\ Para \ y=1 \Rightarrow x = 2\\ \\ \\ S=\{3;\frac{2}{3}),(2,1)\}$