Several methods were used to find log₅ 49 ×log₇ 125. Matthew of Queen Mary's Grammar School, Walsall changed the base to base 10 and Labboid, used a change of base to natural logarithms.

log5 49 = ln49 / ln5 log7 125 = ln125 / ln7.

Hence

log5 49 ×log7 125

= ln49 ln5 × ln125 ln1

= ln(72) ×ln(53) ln5 ×ln7

= 2ln7 ×3ln5 ln5 ×ln7

= 2 ×3 = 6

Patrick, Saul, Hyeyoun, Marc and Arun used this identity log_a b×log_b a = 1 to find the solution as follows. Now log₅ 49=2log₅ 7 and log₇ 125 = 3log₇ 5 therefore

log₅ 49 ×log₇ 125 = 2log₅ 7 ×3log₇ 5 = 6 ×log₅ 7 ×log₇ 5 = 6.

Yatir Halevi generalized this problem to: log_a b^c × log_b a^d. Using the logarithms rules: log_a b^r = r×log_a b and log_b a = 1/log_a b we get the expression equal to cd (log_a b ×log_b a) = cd. So our expression is equal to 2×3=6.