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Question Number 194165 by pascal889 last updated on 29/Jun/23

Answered by qaz last updated on 29/Jun/23

log_2 4096+[4(log_2 x)^3 −16log_2 x]log_2 x=0 log_2 x=y ,12+4y^4 −16y^2 =0 y=±1 ,±(√3) ⇒x=2^(±1) ,2^(±(√3))

log24096+[4(log2x)3−16log2x]log2x=0log2x=y,12+4y4−16y2=0y=±1,±3⇒x=2±1,2±3

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