## College Algebra 7th Edition

$x=1$
We need to solve: $2^{x}-10(2^{-x})+3=0$ We multiply through by $2^x$: $2^x(2^{x}-10(2^{-x})+3=0)$ $2^{2x}-10+3* 2^{x}=0$ $(2^{x}+5)(2^{x}-2)=0$ $(2^{x}+5)=0$ or $(2^{x}-2)=0$ $2^x=-5$ or $2^x=2$ $x=\log_2 -5$=no solution or $x=\log_2 2=1$