1.下列式子中成立的是(假定各式子均有意义)A.logax·logay=loga(x+y) B,(logax)^n=nlogaxC,logax/n=loga n次根号下x D,logax/logay=logax-logay2.已知lg3=a,lg4=b,则log12=()A.b/a+b B.a/a+b C.a+b/a D.a+b/b3.lg4+2lg5=()A.2 B.3 C.1
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 13:32:18
![1.下列式子中成立的是(假定各式子均有意义)A.logax·logay=loga(x+y) B,(logax)^n=nlogaxC,logax/n=loga n次根号下x D,logax/logay=logax-logay2.已知lg3=a,lg4=b,则log12=()A.b/a+b B.a/a+b C.a+b/a D.a+b/b3.lg4+2lg5=()A.2 B.3 C.1](/uploads/image/z/7108870-22-0.jpg?t=1.%E4%B8%8B%E5%88%97%E5%BC%8F%E5%AD%90%E4%B8%AD%E6%88%90%E7%AB%8B%E7%9A%84%E6%98%AF%EF%BC%88%E5%81%87%E5%AE%9A%E5%90%84%E5%BC%8F%E5%AD%90%E5%9D%87%E6%9C%89%E6%84%8F%E4%B9%89%EF%BC%89A.logax%C2%B7logay%3Dloga%28x%2By%29+B%2C%28logax%29%5En%3DnlogaxC%2Clogax%2Fn%3Dloga+n%E6%AC%A1%E6%A0%B9%E5%8F%B7%E4%B8%8Bx+D%2Clogax%2Flogay%3Dlogax-logay2.%E5%B7%B2%E7%9F%A5lg3%3Da%2Clg4%3Db%2C%E5%88%99log12%3D%EF%BC%88%EF%BC%89A.b%2Fa%2Bb+B.a%2Fa%2Bb+C.a%2Bb%2Fa+D.a%2Bb%2Fb3.lg4%2B2lg5%3D%28%29A.2+B.3+C.1)
1.下列式子中成立的是(假定各式子均有意义)A.logax·logay=loga(x+y) B,(logax)^n=nlogaxC,logax/n=loga n次根号下x D,logax/logay=logax-logay2.已知lg3=a,lg4=b,则log12=()A.b/a+b B.a/a+b C.a+b/a D.a+b/b3.lg4+2lg5=()A.2 B.3 C.1
1.下列式子中成立的是(假定各式子均有意义)A.logax·logay=loga(x+y) B,(logax)^n=nlogax
C,logax/n=loga n次根号下x D,logax/logay=logax-logay
2.已知lg3=a,lg4=b,则log12=()
A.b/a+b B.a/a+b C.a+b/a D.a+b/b
3.lg4+2lg5=()
A.2 B.3 C.1 D.-1
4.若lga,lgb是方程2x^2-4x+1=0的两个根,则(lga/b)^2的值等于()
A.2 B.1/2 C.4 D.1/4
5.若log以2为底3的对数=a,则log以2为底27的对数+2log以2为底6的对数可用a表示为__________
6.1/log以1/2为底6的对数+1/log以8为底6的对数+1/log以9为底6的对数=__________
7.求值:(1)log以9为底27的对数 (2)log以8为底9的对数×log以27为底32的对数
(第2题题目中那个log12改成log(4)12
1.下列式子中成立的是(假定各式子均有意义)A.logax·logay=loga(x+y) B,(logax)^n=nlogaxC,logax/n=loga n次根号下x D,logax/logay=logax-logay2.已知lg3=a,lg4=b,则log12=()A.b/a+b B.a/a+b C.a+b/a D.a+b/b3.lg4+2lg5=()A.2 B.3 C.1
1D.了解运算法则即可
2看到你的修改了,用对数换底公式.原式=lg12/lg4=(lg3+lg4)/lg4=(a+b)/b
3A 2lg5=lg25,然后lg4+lg25=lg100=2
4.(lga/b)^2=(lga-lgb)^2=(lga+lgb)^2-4lgalgb(韦达定理)=2^2-4*1/2=2
5.原式=log以2为底(3^3)+2*(log以2为底2+log以2为底3)=3*log以2为底3+2*(log以2为底2+log以2为底3)=3a+2*(1+a)=5a+2
6.原式=1/-log以2为底6 +1/log以2^3为底6 +1/log以3^2为底6=-1/(1+log以2为底3) +3/(log以2为底2+log以2为底3) +2/(log以3为底3 +log以3为底2) 你把log以2为底3设为a,log以3为底2就是1/a,消起来方便点
所以原式=2/(1+a) + 2/((a+1)/a)=2/(1+a) +2a/(a+1) =(2a+2)/(a+1)=2
7.(1)=log以3^2为底3^3=1/2 * 3 *log以3为底3 =1.5
(2)=log以2^3为底3^2 * log以3^3为底2^5 =1/3 *2*log以2为底3 * 1/3 * 5 * log以3为底2 =10/9
有任何问题欢迎提问