若logx表示log10x(10为底数),且log(5x^2)=a,则log(4/x^4)=? 求过程!!!!!!!!!!!!
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解由log(5x^2)=a,
得log5+log(x^2)=a
即log(x^2)=a-log5
故log(4/x^4)
=log(4)-log(x^4)
=2log(2)-2log(x^2)
=2log(2)-2(a-log5)
=2log(2)+2log5-2a
=2(log(2)+log5)-2a
=2log(2*5)-2a
=2log(10)-2a
=2-2a
得log5+log(x^2)=a
即log(x^2)=a-log5
故log(4/x^4)
=log(4)-log(x^4)
=2log(2)-2log(x^2)
=2log(2)-2(a-log5)
=2log(2)+2log5-2a
=2(log(2)+log5)-2a
=2log(2*5)-2a
=2log(10)-2a
=2-2a
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