Hi
i have tried to solve these equations but still cannot get the correct answer:
The following is what i have done:
make
then:
has no solution but answer says it is does.
P.S
" 2y^2-20y+64=0 " is incorrect.
2^(2x)-20(2^x)=-64
" Let 2^x be y " is the right direction, but-
Your mistake: 2^(2x) = 2(2^x)^2.
2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )
As for the rest of the steps, I believe you know them already