Me ajudem com essa equação exponencial?49.... Pergunta de ideia deWallisson18

April 2020 1 108 Report

Me ajudem com essa equação exponencial?

49 $49 ^{x } - 8 . 7^{x } + 7 = 0$

Lista de comentários

ArthurPDC $49^x-8\cdot7^x+7=0\Longrightarrow(7^2)^x-8\cdot7^x+7=0\Longrightarrow(7^x)^2-8\cdot7^x+7=0\\\\
p/\;\;7^x=y:\\\\
y^2-8y+7=0\\\\
\Delta=b^2-4\cdot a\cdot c\\
\Delta=(-8)^2-4\cdot1\cdot7\\
\Delta=64-28\\
\Delta=36\\\\
y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{8\pm\sqrt{36}}{2\cdot1}=\dfrac{8\pm6}{2}=4\pm3\Longrightarrow\begin{cases}y_1=7\\y_2=1\end{cases}$

$p/\;\;y=1:\\
7^x=1\Longrightarrow7^x=7^0\Longrightarrow \boxed{x=0}\\\\
p/\;\;y=7:\\
7^x=7\Longrightarrow7^x=7^1\Longrightarrow\boxed{x=1}\\\\\\
\boxed{S=\{0,1\}}$

2 votes Thanks 0