Several methods were used to find $\log_5 49 \times \log_7 125$. Matthew of Queen Mary's Grammar School, Walsall changed the base to base 10 and Labboid used a change of base to natural logarithms.

$$\eqalign{ \log_5 49 &= \ln 49 / \ln 5 \cr \log_7 125 &= \ln 125 / \ln 7.}$$

Hence

$${ {\log_5 49} \times {\log_7 125} }$$

$${ = {{\ln 49} \over {\ln 5}} \times {{\ln {125}} \over {\ln 1}} }$$

$$ = {\ln(7^2) \times \ln(5^3)\over \ln5 \times \ln 7} $$

$$ = {2\ln 7 \times 3\ln5 \over \ln 5 \times \ln 7} $$

$$ = {2 \times 3 = 6} $$

Patrick, Saul, Hyeyoun, Marc and Arun used this identity $\log_a b\times \log_b a = 1$ to find the solution as follows. Now $\log_5 49=2\log_5 7$ and $\log_7 125 = 3\log_7 5$ therefore

$\log_5 49 \times \log_7 125 = 2\log_5 7 \times 3\log_7 5 = 6 \times \log_5 7 \times \log_7 5 = 6.$

Yatir Halevi generalized this problem to: $\log_a b^c \times \log_b a^d$. Using the logarithms rules:

$\log_a b^r= r\times\log_a b$ and $\log_b a = 1/\log_a b$ we get the expression equal to $cd (\log_a b \times \log_b a) = cd$. So our expression is equal to $2\times 3=6$.