Question

Can anyone solve this radical equation?

Answer 1

`4^x-6*2^(x+1)+32=0\\\\\left(2^2\right)^x-6*2^x*2+32=0\\\\\left(2^x\right)^2-12*2^x+32=0\\\\subtitute\ 2^x=t > 0\\\\t^2-12t+32=0\\\\t^2-4t-8t+32=0\\\\t(t-4)-8(t-4)=0\\\\(t-4)(t-8)=0\iff t-4=0\ or\ t-8=0\\\\t=4\ or\ t=8\\\\t=2^x,\ therefore:\\\\2^x=4\ or\ 2^x=8\\\\2^x=2^2\ or\ 2^x=2^3\\\\\boxed{x=2\ or\ x=3}`

Answer 2

`4^x-6\cdot2^(x+1)+32=0\\ (2^x)^2-6\cdot2^x\cdot2+32=0\\ (2^x)^2-12\cdot2^x+32=0\\ (2^x)^2-4\cdot2^x-8\cdot2^x+32=0\\ 2^x(2^x-4)-8(2^x-4)=0\\ (2^x-8)(2^x-4)=0\\\\ 2^x-8=0\\ 2^x=8\\ 2^x=2^3\\ x=3\\\\ 2^x-4=0\\ 2^x=4\\ 2^x=2^2\\ x=2\\\\ \boxed{x=3 \vee x=2}`