求教一道数学对数题(lg2)²+lg2×lg5+lg50+log27 9
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求教一道数学对数题(lg2)²+lg2×lg5+lg50+log27 9
求教一道数学对数题
(lg2)²+lg2×lg5+lg50+log27 9
求教一道数学对数题(lg2)²+lg2×lg5+lg50+log27 9
log(a,b)表示以a为底的b的对数.
所谓的换底公式就是log(a,b)=log(n,b)/log(n,a).
(lg2)²+lg2×lg5+lg50+log(27,9 )
= lg2(lg2+lg5)+lg50+log(3^3,3^2)
=lg2*lg10+lg50+log(3,3^2) /log(3,3^3) 换底公式
=lg2+lg50+2/3
=lg100+2/3
=2+2/3
=8/3
(lg2)²+lg2×lg5+lg50+log27 9
=lg2*(lg2+lg5)+lg50+2/3
=lg2*1+lg(25*2)+2/3
=lg(2*2*25)+2/3
=2+2/3
(lg2)²+lg2×lg5+lg50+log27 9
=lg2(lg2+lg5)+lg50+log27 9
=lg2*lg10+lg50+log27 9
=lg2+lg50+log27 9
=lg100+log27 9
=2+(lg9)/(lg27)
=2+(2lg3)/(3lg3)
=2+2/3
(lg2)²+lg2×lg5+lg50+log27 9
=lg2×lg2+lg2×lg5+lg(5×10)+log3^3 3^2
=lg2×(lg2+lg5)+lg(5×10)+2/3
=lg2×lg10+lg5+lg10+2/3
=lg2+lg5+1+2/3
=lg(2×5)+1+2/3
=1+1+2/3
=8/3
公式:lgA+lgB=lg(A×B)
logn^a n^b=b/a