For x greater than 1, find the least possible value of the expression 2 log10 x - log_x (1 / 100).

Introduction / Context:
This question tests the ability to work with logarithms that have different bases and to analyze a function in order to find its minimum value for x greater than 1. To solve it efficiently, you must use the change of base formula for logarithms and then apply basic calculus or inequality ideas to minimize a simple one variable expression. It is a standard type of aptitude question that mixes algebraic manipulation with an understanding of how logarithmic functions behave.

Given Data / Assumptions:
- The variable x satisfies x greater than 1.
- The expression to be minimized is E(x) = 2 log10 x - log_x (1 / 100).
- All logarithms are common logarithms (base 10) unless another base is explicitly written.
- The expression is defined for x greater than 1 because log_x (1 / 100) requires x positive and not equal to 1.

Concept / Approach:
We first convert the term log_x (1 / 100) into a base 10 logarithm using the change of base formula log_x y = log10 y / log10 x. Then we express the entire function E(x) in terms of t = log10 x, which is positive for x greater than 1. This converts the problem into minimizing a simple algebraic expression in t of the form 2t + 2 / t. That expression can be minimized using basic derivative ideas or the inequality between arithmetic and geometric means.

Step-by-Step Solution:
Let E(x) = 2 log10 x - log_x (1 / 100). Use change of base: log_x (1 / 100) = log10 (1 / 100) / log10 x. Note that 1 / 100 = 10^-2, so log10 (1 / 100) = -2. Therefore log_x (1 / 100) = -2 / log10 x. So E(x) = 2 log10 x - ( -2 / log10 x ) = 2 log10 x + 2 / log10 x. Let t = log10 x. Since x greater than 1, t is greater than 0. Then E(t) = 2t + 2 / t for t greater than 0. Differentiate with respect to t: E'(t) can be used to confirm a minimum but first find where E'(t) is positive. Compute the first derivative: E'(t) is not needed; instead, dE/dt = 2 - 2 / t^2. Set dE/dt = 0: 2 - 2 / t^2 = 0 implies 2 = 2 / t^2, so t^2 = 1. Because t is greater than 0, t = 1. Substitute back: E_min = 2 * 1 + 2 / 1 = 2 + 2 = 4.

Verification / Alternative check:
We can also use the arithmetic mean and geometric mean inequality. For t greater than 0, the arithmetic mean of the two positive numbers 2t and 2 / t is at least twice the geometric mean. That is, (2t + 2 / t) / 2 is greater than or equal to sqrt( (2t) * (2 / t) ) = 2. Hence 2t + 2 / t is greater than or equal to 4, and equality holds when 2t = 2 / t, so t^2 = 1 and t = 1. This confirms that the least value of the expression is 4.

Why Other Options Are Wrong:
2: This would violate the inequality 2t + 2 / t greater than or equal to 4 for t greater than 0.
3: This is less than 4 and therefore cannot be the minimum value.
5: This is larger than the minimum 4; it is simply another attainable value for some other x, not the least value.

Common Pitfalls:
A common error is to forget that log_x (1 / 100) is negative and to write it as 2 / log10 x instead of minus 2 / log10 x. Another mistake is to try to minimize the original expression directly in terms of x, which is more complex than working with t = log10 x. Some students also confuse the formula for the number of digits with this minimization problem, because both use log10. Carefully converting to a single variable and using either differentiation or the arithmetic mean and geometric mean inequality avoids these problems.

Final Answer:
The least possible value of the given expression is 4.