Question

Root of the equation 3^(x-1)+3^(1-x)=2 is:

Answer 1

$3^{x-1}+3^{1-x}=2\\3^{x-1}+3^{-(x-1)}=2\\3^{x-1}+(3^{x-1})^{-1}=2\\\\t=3^{x-1}\\t+t^{-1}=2\\t+\frac1t=2\\For\ all\ y:\ 3^y>0,\ so\ t>0\ and\ we\ can\ multiply\ both\ sides\ by\ t\\t^2+1-2t=0\\(t-1)^2=0\\t=1\\\\t=3^{x-1}=1\\3^{x-1}=3^0\\x-1=0\\x=1$
Check:
$3^{x-1}+3^{1-x}=3^0+3^0=2\\Left=Right$