高手请进,几道指对方程～～(!)x^(lgx)=x^3/100(2)f(x)=lg(ax+1)在(-无穷大,1)上有意义,a的取值范围(3)(lg)^x-2lgx-2=0有两个根a,b,则loga(b)+logb(a)=.(4)若a>0,2^(3x)-2^x=a(2^x-2^(-x))的解(5)5^(x-1)=3^(x^-1)(6)log5(4^x+144)

Question

高手请进,几道指对方程～～
(!)x^(lgx)=x^3/100
(2)f(x)=lg(ax+1)在(-无穷大,1)上有意义,a的取值范围
(3)(lg)^x-2lgx-2=0有两个根a,b,则loga(b)+logb(a)=____.
(4)若a>0,2^(3x)-2^x=a(2^x-2^(-x))的解
(5)5^(x-1)=3^(x^-1)
(6)log5(4^x+144)-4log5(2)=log[5(2^(x-2)+1)]
单独的^为平方.
谢!
要详细过程～～

Answer 1

1.
x^(lgx)=x^3/100
10^((lgx)^2)=10^(3lgx - 2)
这里 x 应该是实数 ..
所以..
(lgx)^2 = 3lgx - 2
(lgx)^2 - 3lgx + 2 = 0
lgx1 = 1,lgx2 = 2
x1 = 10,x2 = 100
2.
由题意
x>=1 ax+1= 1
a 0
所以
k1 = sqrt(a),k2 = -sqrt(a),k3 = 1
所以
x1 = log(2)sqrt(a),..,x3 = 0
5.
5^(x-1)=3^(x^2-1)
lg5(x-1)=lg3(x-1)(x+1)
(x-1)(lg5-xlg3-lg3) = 0
x1 = 1,x2 = (lg5-lg3)/lg3
6.
log5 (4^x+144)-4log5 (2)=log5 [5(2^(x-2)+1)]
(4^x+144)/(2^4) = 5(2^(x-2)+1)
2^(2x)/16 + 144/16 = 5 * 2^(x)/4 + 5
2^(2x)/16 - 5 * 2^(x)/4 + 9 - 5 = 0
2^(2x) - 20 * 2^x + 64 = 0
(2^x - 16)(2^x - 4) = 0
x1 = 4,x2 = 2