lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/17 08:08:40
![lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c](/uploads/image/z/14335271-71-1.jpg?t=lg%5Ba%5Elga%29%2Blg%28b%5Elgb%29%2Blg%28c%5Elgc%29%E4%B8%BA%E4%BB%80%E4%B9%88%E7%AD%89%E4%BA%8Elg%26%23178%3Ba%2Blg%26%23178%3Bb%2Blg%26%23178%3Bc)
x)INIOIHd +YɎ]Ov7<|m]}9jʆ։0VlTOElkgVN~zBY-,|6mm.DJL>ؒ Tr-P8<Dm/.H̳9 L
lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c
lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c
lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c
公式:loga b的m次方=mloga b
∴lg a 的(lga)=lga ×lga =lg²a
lga-lgb=?(lga)/(lgb)=?lg(a/b)=?
lga-lgb=lg( a / b)lga+lgb=lg( a X b)lga X lgb=?lga / lgb=?
lg(a+b)=lga+lgb?lg(a—b)=lga-lgb?判断题
是lg(ab)=lga+lgb还是lg(a+b)=lga·lgb?
lgA乘以lgB=lg(A+B),那么lgA+lgB会不会=lg(AB)
lg[a^lga)+lg(b^lgb)+lg(c^lgc)为什么等于lg²a+lg²b+lg²c
lg(a+b)/2+lg(b+c)/2+lg(a+c)/2>lga+lgb+lgc
求证lg(a+b)/2+lg(b+c)/2+lg(c+a)/2>lga+lgb+lgc
根号(lga+lgb),1/2(lga+lgb),lg(a+b/2),比较大小
若lg(a-b)+lg(a+b)=lg2+lga+lgb,求a/b的值
若lg(a-b)+lg(a+b)=lg2+lga+lgb,求a/b的值
高中数学求证:a^lgb*b^lga=1还有一个问题:lg(a^lgb)=lga*lgb ?
关于lgb/lga+lga/lgb=5/2如何得到 2lg^2b-5lga*lgb+2lg^2a=0的呀?
已知a>4,b>1,且lg(a+b)=lga+lgb,则lg(a-1)+lg(b-1)=?
已知a>4,b>1,且lg(a+b)=lga+lgb,则lg(a-1)+lg(b-1)=?
证:lg(a+b/2)+lg(b+c/2)+lg(c+a/2)>lga+lgb+lgc,abc不全相等
若lgA=a,lgB=b,lgC=c,则(lgA^2)*[lg(B/2]*(lg√C)=?
已知ab>0,为什么lg(ab)不等于lga+lgb;lga/b不等于lga-lgb