Question

resolva as equações exponenciais 9^x - 10 . 3^x + 9 = 0​

Answer 1

$\large\boxed{\begin{array}{l}\sf 9^x-10\cdot3^x+9=0\\\sf (3^x)^2-10\cdot 3^x+9=0\\\sf a=1~~b=-10~~c=9\\\sf\Delta=b^2-4ac\\\sf\Delta=(-10)^2-4\cdot1\cdot9\\\sf\Delta=100-36\\\sf\Delta=64\\\sf 3^x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf 3^x=\dfrac{-(-10)\pm\sqrt{64}}{2\cdot1}\\\\\sf 3^x=\dfrac{10\pm8}{2}\begin{cases}\sf 3^x=\dfrac{10+8}{2}=\dfrac{18}{2}=9\\\sf 3^x=3^2\\\sf x=2\\\sf 3^x=\dfrac{10-8}{2}=\dfrac{2}{2}=1\\\sf 3^x=3^0\\\sf x=0\end{cases}\\\sf S=\{2,0\}\end{array}}$