Several methods were used to find $\log_{5} 49 \times \log_{7} 125$ . Matthew Poulton of Queen Mary's Grammar School, Walsall changed the base to base 10 and someone, only identifying him or herself as Labboid, used a change of base to natural logarithms.

$\begin{matrix} \log_{5} 49 & = \ln 49 / \ln 5 \\ \log_{7} 125 & = \ln 125 / \ln 7 . \end{matrix}$

Hence

$\log_{5} 49 \times \log_{7} 125$

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

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

$= \frac{2 \ln 7 \times 3 \ln 5}{\ln 5 \times \ln 7}$

$= 2 \times 3 = 6$

Patrick Snow, Saul Forresta, Hyeyoun Chung, Marc Carlambe and Arun Iyer, used this identity 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 $2 \times 3 = 6$ .