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