作业帮log9为底4的对数
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 19:40:58
1、换底公式原式=(lg3/lg4)(lg2/lg9)+log2(4)+log2(√35)=(lg3/2lg2)(lg2/2lg3)+2+log2(√35)=1/4+2+log2(√35)=9/4+l
(log43+log83)(log32+log92)=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)=
log2√2+log927+4^log413=log22^(1/2)+log(3^2)(3^3)+13=1/2+3/2+13=15
log(2^n)(3^n)=n/nlog23=log23log(9)8^(1/n)=1/n*log98原式=(log23+log23+...+log23)*1/n*log98=nlog23*1/n*l
log以5为底4的对数xlog以27为底125的对数=log54xlog27125=lg4/lg5x(lg125/lg27)=lg4/lg5x(3lg5/3lg3)=lg4/lg3=log34
log9((⁴√27)/3)+lg20+log10025+5^(log52)=1/2log₃(⁴√27/3)+lg20+2/2*lg5+2=1/2[log̀
(log94^更号27)/3+lg20+log10025+5^(log52)=(log99^3/8)/3+1+lg2+lg5+2=3/8÷3+1+1+2=1/8+4=4又1/8
a=lg2/lg9=lg2/2lg3所以lg2/lg3=2ab=lg5/lg(1/3)=(1-lg2)/(-lg3)=lg2/lg3-1/lg3=2a-1/lg3所以1/lg3=2a-b原式=lg12
5的2倍log以5为底3的对数这个是5的2倍log以5为底3的对数的次方吗?如果是的话,原式=4log3²（2）-[log3(32/9)-log3(8)]-5^log5(3²）=（
答：3^{log9[(2+√3)^2]}因为：log9[(2+√3)^2]=2log9(2+√3)=2log3(2+√3)/log3(9)=log3(2+√3)所以：3^{log9[(2+√3)^2]
用换底公式log(a)b=[log(c)b]/[log(c)a]就可以化简了将上式都换成以10为底的对数就行了log4³×log9²+后面的看不懂（期望追问）log4³×
log3^3/log2^3xlog5^2/log3^2xlog2^4/log5=lg3/lg2xlg5/lg3x4lg2/lg5(先用公式把指数去掉,然后用换底公式）=lg3/lg2xlg5/lg3x
(log2x)*(log2x/4)=3(log2x)*(log2x-log24)=3(log2x)*(log2x-2)=3设log2x=t则t*(t-2)=3t^2-2t-3=0(t-3)(t+1)=
log34与log43互为倒数log43=1/log34=log34的负一次=-log34log34+log43=log34-log34=0最后答案为0
(以2为底3的对数+以4为底9的对数+以8为底27的对数+以16为底81的对数+以32为底243的对数）-5倍的以2为底二分之三的对数=lg3/gl2+2lg3/(2lg2)+3lg3/(3lg2)+
-5log9(4)+log3(32/9)-5^[log5(3)]-(1/64)^(-2/3)=-5[log3(4)]/log3(9)]+log3(32)-log3(9)-3-64^(2/3)=-5lo