解下列不等式:3·log₃(log₃x) +log ‹1/3› [log₃(9·³√x)]≥1
3·log₃(log₃x) +log ‹1/3›[log₃9+log₃∛x]≥1
3·log₃(log₃x) +log ‹1/3›[2+(1/3)log₃x]≥1
3log‹1/3›[1/log₃x]+log‹1/3›[2+(1/3)log₃x]≥1【这里是把第一个对数的底数和真数都取倒数】
log‹1/3›{[1/log₃x]³[2+(1/3)log₃x]}≥1
故0
解下列不等式:3·log₃(log₃x) +log ‹1/3› [log₃(9·³√x)]≥1
3·log₃(log₃x) +log ‹1/3›[log₃9+log₃∛x]≥1
3·log₃(log₃x) +log ‹1/3›[2+(1/3)log₃x]≥1
3log‹1/3›[1/log₃x]+log‹1/3›[2+(1/3)log₃x]≥1【这里是把第一个对数的底数和真数都取倒数】
log‹1/3›{[1/log₃x]³[2+(1/3)log₃x]}≥1
故0