If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, find the value of x.

Introduction / Context:
This question involves solving an equation that contains common logarithms with base 10. It checks understanding of standard log properties, especially the product rule and the interpretation of the constant 1 as log10 10. The unknown appears inside the arguments of the logarithms, so we must simplify the equation using log rules and then solve a resulting algebraic equation in x. This type of manipulation is very common in aptitude and competitive exams.

Given Data / Assumptions:
- The equation is log10 5 + log10 (5x + 1) = log10 (x + 5) + 1.
- All logarithms are common logarithms (base 10).
- The arguments of the logs must be positive, so 5x + 1 greater than 0 and x + 5 greater than 0, which implies x greater than -1/5 and x greater than -5.
- Because we are interested in a real solution consistent with the answer options, we focus on positive x that satisfy these conditions.

Concept / Approach:
We use two main log identities: log10 A + log10 B = log10 (A * B) and 1 = log10 10. The presence of the constant 1 on the right side can be replaced with log10 10, so that every term in the equation becomes a log10 of some positive quantity. After that, we can combine logs on each side, compare their arguments, and obtain a simple linear equation in x. Solving that gives the required value.

Step-by-Step Solution:
Start with log10 5 + log10 (5x + 1) = log10 (x + 5) + 1. Rewrite 1 as log10 10, because log10 10 = 1. The equation becomes log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10. Use the product rule on the left: log10 5 + log10 (5x + 1) = log10 [5(5x + 1)]. Use the product rule on the right: log10 (x + 5) + log10 10 = log10 [10(x + 5)]. So we have log10 [5(5x + 1)] = log10 [10(x + 5)]. For positive arguments, equality of logs with the same base implies equality of the arguments. Therefore 5(5x + 1) = 10(x + 5). Expand: 25x + 5 = 10x + 50. Rearrange: 25x - 10x = 50 - 5, so 15x = 45. Hence x = 45 / 15 = 3.

Verification / Alternative check:
Substitute x = 3 back into the original equation. The left side is log10 5 + log10 (5(3) + 1) = log10 5 + log10 16. The right side is log10 (3 + 5) + 1 = log10 8 + 1. Numerically, log10 5 + log10 16 = log10 80, while log10 8 + 1 = log10 8 + log10 10 = log10 80. Both sides are equal, which confirms that x = 3 satisfies the equation. The domain conditions are also satisfied because 5x + 1 and x + 5 are positive when x = 3.

Why Other Options Are Wrong:
1: If x = 1, then 5x + 1 = 6 and x + 5 = 6, giving log10 5 + log10 6 on the left and log10 6 + 1 on the right. These are not equal.
5: If x = 5, then 5x + 1 = 26 and x + 5 = 10, which leads to unequal expressions after applying the log rules.
10: If x = 10, then the expressions involve large and unequal arguments, which again do not satisfy the equation when checked numerically.

Common Pitfalls:
A typical mistake is forgetting to interpret the constant 1 as log10 10, causing students to attempt unnecessary operations like exponentiating both sides too early. Another error is combining logs incorrectly, for example treating log10 (x + 5) + 1 as log10 (x + 6) instead of log10 (10(x + 5)). Students also sometimes forget the domain restriction that arguments of logarithms must be positive, leading to extraneous solutions in more complex cases. Using the log product rule systematically and equating arguments only after combining logs on each side helps avoid these issues.

Final Answer:
The value of x that satisfies the equation is x = 3.