# represent $\log_{35}(28)$ by $\log_{14}(7)$ and $\log_{14}(5)$

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.

• $\log_{14}28=2-\log_{14}7$ – J. W. Tanner Sep 3 at 12:53

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 $$\log_{14}(28\cdot7)=\log_{14}(14\cdot14)=2$$.

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

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

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

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