Simplifying two logarithms with different bases I am being asked to simplify:
$(\log_4 7)(\log_7 5)$
How can this be simplified given that the bases are different?
 A: Here's a way that may be the easiest to understand, using the change-of-base formula in its simplest form:
$$
(\log_4 7)(\log_7 5) = \frac{\log_e 7}{\log_e 4} \cdot \frac{\log_e 5}{\log_e 7} = \frac{\log_e 5}{\log_e 4} = \log_4 5.
$$
Here's a way that uses a corollary of the change-of-base formula:
$$
\underbrace{(\log_4 7)(\log_7 5) = (\log_7 7)(\log_4 5)}_\text{a corollary of the change-of-base formula} = 1\cdot\log_4 5. 
$$
Here's the corollary:
$$
(\log_A P)(\log_B Q)(\log_C R)(\log_D S)\cdots = (\log_D P)(\log_A Q)(\log_B R)(\log_C S)\cdots
$$
and generally you can permute the subscripts $A,B,C,D,\ldots$ in any way at all while leaving the arguments $P,Q,R,S,\ldots$ where they are, without changing the value of the product.
That follows from the change-of-base-formula, which is actually a special case of it.
A: use $\log_a b=‎\dfrac{\log b}{\log a}$
A: If in doubt convert to powers so that $$7=4^a; 5=7^b=(4^a)^b=4^{ab}$$ and then extract $ab$ from this equation.
A: Two useful identities

*

*$\log_a b = \dfrac{1}{\log_b a}$

*$log_a b = \dfrac{\log_c b}{\log_c a}$
\begin{align}
   \log_4 7 \cdot \log_7 5 
   &= \dfrac{\log_7 5}{\log_7 4} \\
   &= \log_4 5
\end{align}
