Question

Equação exponencial.

(FGV-SP)se x é a raiz da equação, entao determine $x^{-1}$



$3^{x-1} +3^{x}+3^{x+1}= \frac{13}{27}$

Answer 1

$3^x\cdot3^{-1}+3^x+3^x\cdot3^1=\frac{13}{27}\\\\\frac{3^x}{3^1}+3^x+3\cdot3^x=\frac{13}{27}$

 Consideremos $3^x=k$, segue,

$\frac{k}{3}+k+3k=\frac{13}{27}\\\\\frac{k}{3/9}+\frac{4k}{1/27}=\frac{13}{27/1}\\\\9k+108k=13\\\\117k=13\\\\k=\frac{13}{117}\\\\k=\frac{1}{9}$

 Portanto,

$3^x=k\\\\3^x=\frac{1}{9}\\\\3^x=\frac{1}{3^2}\\\\3^x=3^{-2}\\\\x=-2\\\\x^{-1}=(-2)^{-1}\\\\\boxed{\boxed{x^{-1}=-\frac{1}{2}}}$