Question

помогите пожалуйста

Answer 1

$log_{3}( {2}^{1 - x} ) + log_{3}( {4}^{x} + 2) = log_{3}(9)$

2^1-x•(4^x + 2) = 9

2^x+1 + 2^2-x = 9

2•2^x + 4/2^x = 9

2^x = n, n>0

2n²-9n+4 = 0

D = 81-32 = 49

n = 9±7/4

n1 = 4

n2 = 1/2

1)2^x = 4

x = 2

2)2^x = 1/2

x = -1