Question
( log _{2}left(4^{x}+1right)=x+log _{2}left(2^{x+3}-6right) )
( Rightarrow quad log _{2}left(4^{x}+1right)-log _{2}left(8 cdot 2^{x}-6right)=x )
( Rightarrow log _{2}left(frac{4^{x}+1}{8 cdot 2^{x}-6}right)^{2}=x )
( Rightarrow frac{4^{x}+1}{8 cdot 2^{x}-6}=2^{x} Rightarrow 2^{2 x}+1=8 cdot 2^{2 x}-6 cdot 2^{x} )
( Rightarrow quad a_{2} cdot 7 cdot 2^{2 x}-6 cdot 2^{x}-1=0 )
( Rightarrow quad 7 cdot 2^{2 x}-7 cdot 2^{x}+2^{x}-1=0 )
( Rightarrow quad 7 cdot 2^{x}left(2^{x}-1right)+1left(2^{x}-1right)=0 )
( Rightarrowleft(7 cdot 2^{x}+1right)left(2^{x}-1right)=0 )
( Rightarrow quad 2^{x}=frac{-1}{7}, 1 )
( therefore 2^{x} eq frac{-1}{7} Rightarrow 2^{x}=1 )
( Rightarrow quad|x=0| )

5. log2 (4* +1) = x +log, (2*** - 6)
Solution
