Asked by JB
Solve the equation. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)
2^x − 8(2^−x) + 2 = 0
2^x − 8(2^−x) + 2 = 0
Answers
Answered by
Steve
2^(-x) (2^(2x) + 2*2^x - 8) = 0
2^(-x) is never zero, so that means you must have
2^(2x) + 2*2^x - 8 = 0
(2^x+4)(2^x-2)
2^x is never negative, so that means
2^x-2=0
2^x=2
x=1
2^(-x) is never zero, so that means you must have
2^(2x) + 2*2^x - 8 = 0
(2^x+4)(2^x-2)
2^x is never negative, so that means
2^x-2=0
2^x=2
x=1