Log On
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 $2\times 3=6$.