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.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\log }_{5}49}& =\ln 49/\ln 5\\ \multicolumn{1}{c}{{\log }_{7}125}& =\ln 125/\ln 7.\end{array}}$

Hence

${\displaystyle {\log }_{5}49\times {\log }_{7}125}$

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

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

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

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

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 $\mathrm{cd}({\log }_{a}b\times {\log }_{b}a)=\mathrm{cd}$. So our expression is equal to $2\times 3=6$.