Question

rezolvati in multimea numerelor reale ecuatia 2019^x + 2019^(-x)=2

Answer 1

$\displaystyle\\2019^x+2019^{-x}=2 \Longleftrightarrow 2019^{x}+\frac{1}{2019^{x}}=2 \Longleftrightarrow\\\frac{2019^{2x}+1}{2019^{x}}=2 \Longleftrightarrow \frac{2019^{2x}-2\cdot 2019^{x} +1}{2019^{x}}=0 \Longrightarrow \\2019^{2x}-2\cdot 2019^{x}+1=(2019^{x}-1)^2=0 \Longleftrightarrow 2019^{x}=1 \Longleftrightarrow \boxed{x=0}.$