The value of \(\left\{ {\frac{1}{{{{\log }_9}60}} + \frac{1}{{{{\log }_{16}}60}} + \frac{1}{{{{\log }_{25}}60}}} \right\}\) is -

Question

Question

Answer 1

CONCEPT:

By base changing theorem we know that \({\log _b}a = \frac{{{{\log }_x}a}}{{{{\log }_x}b}} = \frac{1}{{{{\log }_a}b}}\).
log _x (x) = 1

CALCULATION:

Given that \(\left\{ {\frac{1}{{{{\log }_9}60}} + \frac{1}{{{{\log }_{16}}60}} + \frac{1}{{{{\log }_{25}}60}}} \right\}\)

By base changing we can write it as -

⇒ log₆₀9 + log6016 +log6025

By product rule of logarithms we can write it again as -

⇒ log₆₀(9 x 16 x 25) = log₆₀(3600)

⇒ log₆₀ (60)² = 2 log₆₀(60) = 2

Therefore, option (3) is the correct answer.