Question

log2(2x+3)+log2(x-1)=4-log2 16/3

Answer 1

Ответ:

log2(2x+3)+log2(x-1)=4-log2(16)/3

log(x-1)/log(2) + log(2x+3)/log(2) = 8/3

x = 1/4(√(25+32x2^2/3) - 1)