Equation lgx + lgx ^ 2 +. + lgx ^ n = n ^ 2 + N, then x=

Equation lgx + lgx ^ 2 +. + lgx ^ n = n ^ 2 + N, then x=

lgx+lgx^2+.+lgx^n
=lgx+2lgx+…… +nlgx
=(1+2+…… +n)lgx
=n(n+1)/2*lgx=n²+n
So lgx = 2
x=10²=100

The interval where the equation X-1 = lgx must have a root is ()
A. （0.1，0.2）B. （0.2，0.3）C. （0.3，0.4）D. （0.4，0.5）

Let f (x) = x-1-lgx, then f (0.1) = 0.1-1-lg0.1 = 0.1 ＞ 0, f (0.2) = 0.2-1-lg0.2 = 0.2-1 - (lg2-1) = 0.2-lg2, ∵ lg20.2 = LG2 & nbsp; 10.2 = LG32 ＞ LG10 = 1; ∵ LG2 ＞ 0.2; f (0.2) ＜ 0; similarly: (0.3) = 0.3-1-lg0.3 - (lg3-1) = 0

The equation lgx-9 / x = 0 has () roots

In this problem, if we make the image of lgx and 9 / X respectively, we can easily know that there is only one zero point, x0, and x0 > 1
To prove, Let f (x) = lgx-9 / X
The domain is x > 0,
f'(x)=1/(xln10)+9/x^2>0
That is to say, f (x) increases monotonically and has only one zero point at most
And f (1) = - 90
So there is a unique zero between (1,10)

How many real roots does the equation x × lgx = 1 have in the interval (2,3)?

The question is simple
The function f (x) = x × lgx-1. It is obvious that the image of the function in [2,3] is a continuous curve, and
F(2)=2lg2-1=lg(2/5)lg1=0
So f (2) f (3)