Question

Me ajudem com essa equação exponencial?

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

Answer 1

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

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