(LOG6 3)^2+(log618)/(log26)
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(LOG63)^2+(log618)/(log26)(LOG63)^2+(log618)/(log26)(LOG63)^2+(log618)/(log26)(log63)^2+(log618)/(lo
(LOG6 3)^2+(log618)/(log26)
(LOG6 3)^2+(log618)/(log26)
(LOG6 3)^2+(log618)/(log26)
(log6 3)^2+(log618)/(log2 6)
=(lg3/lg6)^2+(lg18/lg6)/(lg6/lg2)
=(lg3 / lg6)^2+[ (2lg3+lg2) / lg6) *(lg2 / lg6)
=(lg3)^2+(2lg3lg2/+(lg2 )^2 / (lg6)^2
=(lg3+lg2))^2 / (lg6)^2
=1
(LOG6 3)^2+(log618)/(log26)
((1-log6(3))²+log6(2)*log6(18))/log6(4)
[(1-log6 3)²+log6 2×log6 18]/log6 4
(log6 2)^2+log6 3*log6 12
(log6^2)^2+(log6^3)*(log6^4)+(log6^3)^2
(log6 3)^2+log6 2+log6 2log6 3=?
[(1-log6 3)^2+log6 2-log6 18]/log6 4
2log6(3)-log6(54)等于?
〔(1-log63)的平方+log62乘以log618〕÷log64(log6(2))^2这步是怎么算出来的,
[(1-log6底3)^2+log6底2*log6底18]÷log6底4=
计算:log6(3)×log6(12)+(log6(2))²
求值:(log6 2)²+log6 3*log 6 2+log6 3,应该怎么算?
(log6^4)(log6^18)+2(log6^3)^2
(log6 3)^2+log6 18乘以log6 2急求!!!!!!!!!!!!!!
求值:log6^3+log6^2=log6^6为什么?
log6 2=a log6 3=?
数学〔(1-log63)的平方+log62乘以log618〕÷log64(log6(2))^2这步是怎么算出来的 6是底数
求值:[log6(3)]^2+log6(18)/log2(6)很着急.我已经知道演算的过程,想知道第三部为什么+[log6(2)]^2要有(乘)变为(加),而且为什么要将它平方?[log6(3)]^2+log6(18)/log2(6)=[log6(3)]^2+log6(3^2 * 2)*[log6(2)]=[log6(