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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.

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    $\begingroup$ $\log_{14}28=2-\log_{14}7$ $\endgroup$ – J. W. Tanner Sep 3 at 12:53
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You are correct that $\log_{14}28$ cannot be simplified

by expanding $\log_{14}(7\cdot5-7)\ne\log_{14}(7\cdot5)-\log_{14}5$.

Instead, I would suggest using $\log_{14}28=2-\log_{14}7$.

To see that, note that $\log_{14}(28\cdot7)=\log_{14}(14\cdot14)=2$.

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$$a(\log5+\log7)=2\log2+\log7=2\log2+a\log5+(1-a)\log7=0$$

$$b(\log2+\log7)=\log5\iff b\log2-\log5+b\log7=0$$

Let $c=\log_{14}5,$ $$\implies c\log2-\log5+c\log7=0$$

$$\implies\det\begin{pmatrix} 2 & a & 1-a \\ b & -1 & b \\ c & -1 &c\end{pmatrix}=0$$

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