Question:
If log (2a × 3b × 5C) is the arithmetic mean of log (22 × 33 × 5), log (26 × 3 × 57 ), and log (2 × 32 × 54 ), then a equals
If log (2a × 3b × 5C) is the arithmetic mean of log (22 × 33 × 5), log (26 × 3 × 57 ), and log (2 × 32 × 54 ), then a equals
Updated On: Jul 29, 2025
- 4
- 5
- 3
- 2
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The Correct Option is C
Solution and Explanation
To solve for a, we need to find the arithmetic mean of the three given logarithmic expressions and set it equal to the given expression:
The expression we have is: log(2a × 3b × 5c)
The arithmetic mean of the three expressions is
(log(22 × 33 × 5) + log(26 × 3 × 57) + log(2 × 32 × 54)) / 3
Using the property of logarithms, log(mn) = log(m) + log(n), we can expand each term:
- log(22 × 33 × 5) = log(22) + log(33) + log(5)
- log(26 × 3 × 57) = log(26) + log(3) + log(57)
- log(2 × 32 × 54) = log(2) + log(32) + log(54)
Evaluating the additions:
- First set: log(22) + log(33) + log(5) = 2log(2) + 3log(3) + log(5)
- Second set: log(26) + log(3) + log(57) = 6log(2) + log(3) + 7log(5)
- Third set: log(2) + log(32) + log(54) = log(2) + 2log(3) + 4log(5)
Add and then divide by 3 to find the arithmetic mean:
- Arithmetic mean = [(2log(2) + 3log(3) + log(5)) + (6log(2) + log(3) + 7log(5)) + (log(2) + 2log(3) + 4log(5))] /3
- Simplify: [(2+6+1)log(2) + (3+1+2)log(3) + (1+7+4)log(5)] / 3
- Result: [9log(2) + 6log(3) + 12log(5)] / 3 = 3log(2) + 2log(3) + 4log(5)
Given: log(2a × 3b × 5c) = 3log(2) + 2log(3) + 4log(5)
- Expand: log(2a) + log(3b) + log(5c)
- This implies: alog(2) + blog(3) + clog(5) = 3log(2) + 2log(3) + 4log(5)
Compare coefficients for each base:
- For base 2: a = 3
- For base 3: b = 2
- For base 5: c = 4
Thus, the value of a is 3.
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