计算log8底9*log3底32
人气:251 ℃ 时间:2020-04-07 20:39:09
解答
备注:第一个()内是“底数”,第二个()内是“真数”
log8底9*log3底32 =log(3的平方)(2的三次方)*log(2的五次方)(3)=3/2log(3)(2)*1/5log(2)(3)=3/10
推荐
- 请问这道题怎么算 设log8(9)=a,log3(5)=b,则lg2= (用a,b表示)
- 计算(log4 3+ log8 3)(log3 2+ log9 2)-- log1/2 32^(1/3)+5^(log259)
- log3(2)*log8(9)怎么计算?
- log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)
- (log2(3)+log8(9))x(log3(4)+log9(8)+log3(2))
- Tom has got three jackets.对three提问
- 已知数列{an}满足a1=4,an+1=an+p.3^n+1(n属于N+,P为常数),a1,a2+6,a3成等差数列.
- 一个长方体截成两段,成为两个相同正方体,表面积增加32平方厘米,原来这个长方体体积【 】立方厘米.
猜你喜欢