For `xgt1`, show that: `2log_10x-log_x 0.01ge4`

solve for x: 2log_10x-log_x(0.01)=5

Show that: |log_ba+log_ab|ge2

For x > 1 , the minimum value of 2 log_10(x)-log_x(0.01) is

The least value of the expression 2(log)_(10)x-(log)_x(0.01)dot for x >1 is (a) 10 (b) 2 (c) -0. 01 (d) 4

The least value of the expression 2(log)_(10)x-(log)_x(0. 01),forx >1, is a. 10 b. 2 c. -0.01 d. none of these

The least value of the expression 2(log)_(10)x-(log)_x(0. 01),forx >1, is a. 10 b. 2 c. -0.01 d. none of these

Solve : 6((log)_x2-(log)_4x)+7=0.

If 0ltxlt1 , prove that: log(1+x)+log(1+x^2)+log(1+x^4)+… oo=-log(1-x)

Statement - 1 : If x gt 1 then log_(10)x lt log_(3)x lt log_(e )x lt log_(2)x . Statement - 2 : If 0 lt x lt 1 , then log_(x)a gt log_(x)b implies a lt b .

Solution set of the in equality log_(10^(2)) x-3(log_(10)x)( log_(10)(x-2))+2 log_(10^(2))(x-2) lt 0 , is :