In learning theory, one occasionally encounters inequalities such as

and

The first one is asking: what is the largest x>0 violating (1)? The second one is asking: what is the smallest x>0 satisfying (2)? Since x grows faster than log⁡x, there is a constant C=C(A,B) such that (1)⟹x≤C and equivalently, x≥C⟹(2). The problem is that x and log⁡x appear on both sides of the inequality, making it a transcendental one. Before you break out your Lambert W function, here is a simple way of obtaining the nearly optimal C. The key is the following consequence of log⁡z≤z−1: log⁡x≤αx−log⁡α−1,∀x,α>0. Let us apply this to (1), with α yet unspecified: x≤A+Blog⁡x≤A+B(αx−log⁡α−1), which implies (1−αB)x≤A−Blog⁡α−B. Let us take α=12B. The constant 1/2 is arbitrary; any number in (0,1) will do. We can't have αB>1, though --- since we're about to divide the inequality by (1−αB). Doing so yields x≤2(A+Blog⁡(2B)−B). Conclusion: (1)⟹x≤2(A+Blog⁡(2B)−B), (2)⟸x≥2(A+Blog⁡(2B)−B).