We need to find the real values of a for which f(x)=18x^2 - lnx >= a

a is the minimum value of the function f(x) = 18x^2 - ln x.

At the minimum value of f(x) = 18x^2 - ln x, f'(x) = 0 and f''(x) is positive.

f'(x) = 36x - 1/x = 0

=> 36x^2 - 1 = 0

=> (6x - 1)(6x + 1) = 0

=> x = 1/6 and x = (-1/6)

f''(x) = 72x, this is positive for x = (1/6), so the minimum value lies at x = (1/6)

At x = 1/6, f(x) = 18x^2 - lnx = 18*(1/36) - ln (1/6)

=> 1/2 - ln (1/6)

So the values of a are a <= (1/2 - ln(1/6))

**The required values that a can take are a < (1/2 - ln(1/6))**

The first thing to do is to analyze the monotony of the function f(x). For this reason, we'll differentiate the function:

f'(x) = 36x - 1/x

We'll put f'(x) = 0.

36x - 1/x = 0

(36x^2 - 1)/x = 0

We notice that we have a difference of 2 squares at numerator:

(36x^2 - 1) = (6x - 1)(6x + 1)

f'(x) = 0 if and only ifÂ (6x - 1)(6x + 1) = 0.

6x - 1 = 0 => x = 1/6

6x + 1 = 0

x = -1/6

The derivative is negative, over the range (0;1/6) and it is positive over the range (1/6 ; +infinite).

That means that the function is decreasing over the interval (0;1/6] and it is increasing over the range [1/6 ; +infinite).

So, the point f(1/6) is a local minimum point for the function.

Therefore, f(x)>=f(1/6)>=a

f(1/6) = 18/36 - ln(1/6) = 1/2 + ln 6

So, a =< 1/2 + ln 6

**All real values of "a" are located in the interval (-infinite ; 1/2 + ln6].**