Question

Resolva o sistema:

100^x . √10^y = 10
0,1^x . 0,01^y/2 = 0,01

Answer 1

Olá Maria,

use as propriedades da exponenciação..

$\begin{cases}100^x\cdot\left( \sqrt{10}\right)^y=10\\ (0,1)^x\cdot(0,01)^{ \tfrac{y}{2} }=0,01 \end{cases}\\\\\\ \Rightarrow\begin{cases}(10^2)^x\cdot\left(10^{ \tfrac{y}{2}\right)=10^1\\ \left( \dfrac{1}{10}\right)^x\cdot\left( \dfrac{1}{100}\right)^{ \tfrac{y}{2} }= \dfrac{1}{100} \end{cases}\\\\\\ \Rightarrow\begin{cases}10^{2x}\cdot10^{ \tfrac{y}{2} }=10^1\\ (10^{-1})^x\cdot(10^{-2})^{ \tfrac{y}{2} }=10^{-2}\end{cases}$

$\\\\\\ \Rightarrow\begin{cases}\not10^{2x+ \tfrac{y}{2} }=\not10^1~~(i)\\ \not10^{-x-y}=\not10^{-2}~~(ii)\end{cases}\\\\\\ \Rightarrow\begin{cases}2x+ \dfrac{y}{2}=1~~(i)\\ -x-y=-2~~(ii)~~(multiplica~por~2~e~soma)\end{cases}\\\\\\ \Rightarrow\begin{cases}~~2x+ \dfrac{y}{2}=1~~(i)\\ -2x-2y=-4~~(ii) \end{cases}\\ ~~~~~--------\\ ~~~~~~~~~~~~- \dfrac{3}{2}y=-3\\\\ ~~~~~~~~~~~~~~~~y=-2$

$2x+ \dfrac{y}{2}=1\\\\ 2x+ \dfrac{-2}{~~2}=1\\\\ 2x+(-1)=1\\ 2x-1=1\\ 2x=1+1\\ 2x=2\\\\ x= \dfrac{2}{2}\\\\ x=1$

Portanto a solução do sistema de equações exponenciais acima é:

$\huge\boxed{S_{x,y}=\{(1,-2)\}}$