log(a)(1 x) x的极限
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据题意得:1-2a>03a-1>0=>1/3x)因此可得:1-2a>3a-1=>a
再答:满意请采纳哦,谢谢再问:log以a为底e的对数是怎么推导到log以e为底e的对数除以log以e为底a的对数的呢?是又用到了对数的性质吗我都忘光了求教谢谢再答:嗯,换底公式再问:还有最后一个问题l
symsxpositivelimit(tan(x)^(1/log(x)),0)ans=exp(1)
log(2x-1)-log(x-1)>0log(2x-1)>log(x-1)①当0
x=a^t带入logax+3logxa-logxy=3得到loga^ty=t+3loga^ta-3所以y=a^【t(t+3loga^ta-3)】化简得y=a^(t^2-3t+3)
由log(a)(4+3X-X^2)-log(a)(2X-1)>log(a)(2),可得:log(a)[(4-X)*(1+X)]-log(a)(2X-1)〉log(a)(2)所以log(a)[(4-X)
[log(底数为a)x]=[1\2log(底数为a)b]-[log(底数为a)c][log(底数为a)x]=【log(底数为a)根号b】-[log(底数为a)c][log(底数为a)x]=log(底数
http://hi.baidu.com/zjhz8899/album/item/a87ddbd0728dff139a50270f.html
①limx→n-(x-[x])=n-(n-1)=1②limx→eln(x-1)/(x-e)=[limx→eln(x-1)*limx→e[1/(x-e)]=ln(e-1)*(+∞)=+∞③limx→0+
∵log(a)(m/n)=log(a)(m)-log(a)(n)∴-loga1/x=-(loga1-logax)∵log(a)(1)=0∴-(loga1-logax)=logax明教为您解答,如若满意
x-4>02x-1>0x>4(1)04(2)a>1x-4>2x-1x4,空集综上所述,x>4
是的loga(1/2次方)x=logax^2=2logax
据题意得:1-2a>03a-1>0=>1/3x)因此可得:1-2a>3a-1=>a
∵log(a)(x+y)=√3,∴x+y=a^√3.∵log(a)x=1,∴x=a^1=a.y=(x+y)-x=(a^√3)-a=a(a^(√3-1)-1).两边取对数,log(a)y=log(a)a
a>1,x+1-a>a,x>2a-10
log(a^-1)x=lgx/lga^-1=lgx/(-lga)=-(lgx/lga)=-loga(x)所以互为相反数
log以2为底X的对数的导数为1/[x*ln(2)]x^3的导数为3x^2整个的导数就是3x^2+1/[x*ln(2)]
结果是c[c^x(1+(a/c)^x+(b/c)^x)]^(1/x)=c(1+(a/c)^x+(b/c)^x)^(1/x)=c因为(a/c)^x,(b/c)^x->0而1/x->0所以c*1^0=c
令b=log(a)(x)则x=a^b则f(x)=|b-1|+2|b|当f(x)