If log base 3 of 30 is equal to 1 divided by a and log base 5 of 30 is equal to 1 divided by b, that is log_3 30 = 1 / a and log_5 30 = 1 / b, then find the value of 3 log base 30 of 2 in terms of a and b.

Introduction / Context:
This algebraic logarithm question asks you to express 3 log base 30 of 2 in terms of two parameters a and b that are defined via log_3 30 and log_5 30. It tests the ability to use the change of base formula, to manipulate logarithmic expressions symbolically, and to express one log in terms of others using factorisation of the number 30. Such symbolic log problems are common in algebra and aptitude practice sets.

Given Data / Assumptions:
- log_3 30 = 1 / a.
- log_5 30 = 1 / b.
- All numbers involved (2, 3, 5, 30) are positive and different from 1, so all required logs are defined.
- We must find the expression for 3 log_30 2 in terms of a and b.

Concept / Approach:
We will rewrite each given relation in terms of a single common base, for example base 10 or base e, using the change of base formula log_p q = ln q / ln p. Then we can express ln 3 and ln 5 in terms of ln 30, a and b. Because 30 factorises as 2 * 3 * 5, we can find ln 2 in terms of ln 30, ln 3 and ln 5. Finally we use log_30 2 = ln 2 / ln 30 and multiply by 3 to obtain the desired expression.

Step-by-Step Solution:
From log_3 30 = 1 / a, apply change of base: log_3 30 = ln 30 / ln 3. So ln 30 / ln 3 = 1 / a, which implies a = ln 3 / ln 30. Thus ln 3 = a ln 30. From log_5 30 = 1 / b, apply change of base: log_5 30 = ln 30 / ln 5. So ln 30 / ln 5 = 1 / b, which implies b = ln 5 / ln 30. Thus ln 5 = b ln 30. Since 30 = 2 * 3 * 5, take logs: ln 30 = ln 2 + ln 3 + ln 5. Substitute ln 3 and ln 5 in terms of ln 30: ln 30 = ln 2 + a ln 30 + b ln 30. Rearrange to solve for ln 2: ln 2 = ln 30 - a ln 30 - b ln 30 = (1 - a - b) ln 30. Now log_30 2 = ln 2 / ln 30 = (1 - a - b) ln 30 / ln 30 = 1 - a - b. So 3 log_30 2 = 3 (1 - a - b).

Verification / Alternative check:
We can check consistency by choosing a specific numerical base to compute approximate values. Let ln denote the natural log. Evaluate log_3 30 numerically, find a, and then compute 3 log_30 2 directly. Comparing it with 3 (1 - a - b) after also computing b from log_5 30 shows that the two sides are equal within rounding error. This provides a numerical verification of the algebraic manipulation.

Why Other Options Are Wrong:
3(1 + a + b): This would arise if ln 2 were incorrectly taken as (1 + a + b) ln 30, which is inconsistent with the factorisation equation ln 30 = ln 2 + ln 3 + ln 5.
2(1 - a - b): This has the correct combination 1 - a - b but an incorrect coefficient 2 instead of 3. The factor of 3 comes directly from 3 log_30 2.
3(1 + a - b): This corresponds to an incorrect expression for ln 2 where ln 5 is subtracted rather than added. That contradicts the identity ln 30 = ln 2 + ln 3 + ln 5.

Common Pitfalls:
A common error is to confuse log_3 30 with log_30 3 and hence invert a and 1 / a incorrectly. Another mistake is to ignore the factor 3 in 3 log_30 2 until the end and then forget to multiply the final result. Students also sometimes mishandle the factorisation 30 = 2 * 3 * 5 and write ln 30 = ln 3 + ln 5 - ln 2, which is wrong. Being systematic with change of base and carefully using the factorisation identity prevents these errors.

Final Answer:
The value of 3 log base 30 of 2 is 3(1 - a - b).