Question

Como resolver?
9^x-28.3^x+27=0

Answer 1

faço : 3^x=y --->  (3^x)²-28.(3^x)=27=0 ====> y²-28y-27=0
a partir disso se resolve a equação e depois para os valores encontrados substitui em 3^x=y! Abraços

Answer 2

Oi ^^

$9^x-28\cdot3^x+27=0~~.\\ (3^2)^x-28\cdot3^x+27=0~~.\\ (3^x)^2-28\cdot3^x+27=0\\\\ fazendo~3^x=m\\\\ (m)^2-28\cdot(m)+27=0\\ m^2-28m+27=0\\\\ \Delta=b^2-4ac\\ \Delta=(-28)^2-4\cdot1\cdot27~~.\\ \Delta=784-108\\ \Delta=676\\\\ m= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-28)\pm \sqrt{676} }{2\cdot1}= \dfrac{28\pm26}{2}\\\\\\ \begin{cases}m'=\dfrac{28-26}{2}= \dfrac{2}{2}=1\\\\ m''=\dfrac{28+26}{2}=\dfrac{54}{2}=27 \end{cases}$

Retomando a variável original, 3^x=m:

$3^x=1~~~~~~~~~~~~~~~~~~~~~~~~3^x=27\\ 3^x=3^0~~~~~~~~~~~~~~~~~~~~~~~3^x=3^3\\ \not3^x=\not3^0~~~~~~~~~~~~~~~~~~~~~~\not3^x=\not3^3\\\\ x'=0~~~~~~~~~~~~~~~~~~~~~~~~~x''=3\\\\\\ \Large\boxed{\boxed{\boxed{S=\{0,3\}}}}.\\.$

Ótimos estudos ;D