Question

20. Give the values of x for which x > 500 lg x.

Note: lg is log base 2 (not ln).

Answer 1

I regret about the last confusion., Here is your solution

x > 500 log x (*)
is NOT an algebraic inequality

consider that if x = 1  
500 log x = 0 and (*) is true
if 0 < x < 1 we know that log x is negative, therefore
x > 500 log x, if 0 < x ≤ 1

the problem arises AFTER x = 1
let's consider log as base e logarithms

we need to find the value such that  
x = 500 log x

in other words, if
f(x) = x - 500 log x
we are looking for a root of  
f(x) = 0

it's easy and quick the Newton method

start with a guess value
x0 = 1
and define
x1 = x0 - f(x0)/f '(x0)
where f '(x) = 1 - 500/x is the derivative, so

x1 = 1 - (1 - 500 log 1)/(1 - 500/1) = 500/499 ≈ 1.002004008

and then recursively
x_n+1 = xn - f(xn)/f '(xn)

until the wanted precision p is reached
|x_n+1 - xn| < p

suppose we want 3 correct decimal digits
define
x2 = x1 - f(x1)/f '(x1) = 1.002 - f(1.002)/f '(1.002) ≈ 1.002006021

since
|x2 - x1| ≈ 2.0·10^(-6) < 10^(-3)

therefore
x = 1.002
is appraximately the value where
α = 500 log α

in conclusion the interval where the given inequality is true is
0 < x < α