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若log3(2次幂)=t ,则log3(8次幂)-log3(108次幂)=多少
人气:297 ℃ 时间:2020-06-04 15:52:21
解答
log3(8次幂)-log3(108次幂)
=log3(2^3)-log3(3^3*2^2)
=3log3(2)-[3log3(3)+2log3(2)]
=3t-3-2t
=t-3
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