Question

log2(9-2^x)=3-x

Answer 1

$log_{2}(9-2^x)=3-x\ 9-2^x=2^{3-x}\ 9-2^x=frac{8}{2^x}\ 9*2^x-2^{2x}=8\ 2^x=t\ -t^2+9t-8=0\ t^2-9t+8=0 \ D=81-4*1*8= 7^2\ t=frac{9+7}{2}=8\ t=fraC{9-7}{2}=1\ 2^x=8\ x=3\ 2^x=1\ x=0$