I'm trying to figure out how to express $\log_{35}(28)$ with $a:=\log_{14}(7)$ and $b:=\log_{14}(5)$ (the hint convert the base to 14 was given).
So, $\log_{35}(28) = \dfrac{\log_{14}(28)}{\log_{14}(35)}$.
I already figured out the denominator is $a+b = \log_{14}(7)+\log_{14}(5) = \log_{14}(5\cdot 7) = \log_{14}(35) \Longrightarrow \log_{35}(28) = \dfrac{\log_{14}(28)}{a+b}$.
But I can't figure out the numerator. My guess is that $7\cdot 5 -\textbf{7}=28$ but there's no rule by which I can perform a subtraction in the argument of the log.
My other guess would be to find something like $x\log_{14}(7)+y\log_{14}(5) =\log_{14}(7^x\cdot 5^y)$ or $x\log_{14}(7)-y\log_{14}(5) =\log_{14}\left(\dfrac{7^x}{5^y}\right)$ so that $7^x\cdot 5^y=28$ or $\dfrac{7^x}{5^y}=28$.
However, I believe that there must be an easier way.