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

• A. 4
• B. 5
• C. 3
• D. 2

Answer 1

The Correct Option is C

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.