Решите уравнение ㏒₂x+㏒₄x+㏒₈x=1
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Ответил ShirokovP
0
log2x + log4x + log8x = 1
log2x + log2^2(x) + log2^3(x) = 1
log2x + 1/2*log2x + 1/3*log2x = 1
log2x (1 + 1/2 + 1/3) = 1
log2x * 11/6 = 1
log2x = 6/11
x = 2^(6/11)
log2x + log2^2(x) + log2^3(x) = 1
log2x + 1/2*log2x + 1/3*log2x = 1
log2x (1 + 1/2 + 1/3) = 1
log2x * 11/6 = 1
log2x = 6/11
x = 2^(6/11)
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