Answers

(1)x^2-x>0 x-1>0であるので
x>1
また
log8(x-1)=log2(x-1)/log2(8)=1/3・log2(x-1)
✴︎に代入して
log2(x)+log2(x-1)-log2(x-1)<a
log2(x)<log2(2^a)
2>0より
x<2^a
全部あわせて
1<x<2^a

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