解方程log2[log3(log4^x)]=0
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x-1>0x>1x+1>0x>-1log3(x-1)+log3(x+1)=1log3(x-1)(x+1)=log33(x-1)(x+1)=3x^2-1=3x^2=4x=2x=-2(舍)
(log34+log38)(log23+log29)+(log3根号2)(log97)/[(log1/37)(log三次根号48)]=(5lg32)(3lg32)+[9(lg23)(lg37)/4]/
其实㏒2^3=lg3/lg2∴log2^3*log3^7*log7^8=(lg3/lg2)*(lg7/lg3)*(lg8/lg7)你在纸上把这列式子写下来发现交叉消去左后结果是=lg8/lg2=log
(log2(5)+log4(125))(log3(2)/log√3(5))=(log2(5)+3/2log2(5))(2log3(2)/log3(5))=5/2log2(5)*2log5(2)=5lo
要求log5跟log3是没有关系的log5的求法是1-log2也就是log10-log2=log10/2=log5log2大约为0.3010所以log5=1-0.3010=0.6990
=lg9/lg2×lg4/lg3=2lg3/lg2×2lg2/lg3=4再问:确定吗?急需再答:嗯再问:有人等于2呀!再答:不信拉倒,再见再问:嗯谢谢了
log2(3)+log3(5)+log3(2)=log2(3)+log3(10)=4.2429387700296再问:=1g3/1g2+(lg5+lg2)lg3这个式子最后那个lg3原来是分母,,怎么
lg2+lg5=lg10=1log2(3)乘log3(2)=1lg2+lg5-log2(3)乘log3(2)=0
log2(3)=1+log2(3/2)log3(4)=1+log3(4/3)0log3(4)>0log2(3)/log3(4)=lg3/lg2/(lg4/lg3)=lg²3/(lg2lg4)
log2(2-x)=log2(x-1)+log(2)log2(2-x)=log2[2(x-1)]2-x=2x-2x=4/3带入检验,成立所以x=4/3
没有错...换底公式的运用于逆运用.log(2)(3)xlog(3)(7)=ln3xln7/ln2xln3=ln7/ln2=log(2)(7)
解题思路:考查换底公式的应用解题过程:最终答案:略
log3(x+2)>0X+2>1X>-1log2(x-3)
呵呵,我不知道你目前是哪个年级的哦,这道题其实有初级简单的解法,也有高级复杂的解法,索性我都跟你讲了吧.首先初级解法:log2(3)=(ln3)/(ln2),同理log3(4)=(ln4)/(ln3)
log2[log3(log4(x))]=0log3(log4(x))=1log4(x)=3x=4^3=64
用两次换底公式:第一次:变为lg3/lg2*lg4/lg3可消去lg3,变为lg4/lg2第二次:把lg4/lg2变为log2^4所以结果就等于2
log3(x+2)>0即log3(x+2)>log3(1)∴x+2>1∴x>-1即不等式解集为(-1,+∞)log2(x-3)
由log2(log3^x)=1知,log3^x=2,从而x=9,所以方程的解集为{x|x=9},这样解集的子集有两个:空集和{x|x=9}本身.此解答仅供参考.
log33.14>log33=1log20.8log20.8