Via @markritchings, an excellent logs problem:

If $a = \log_{14}(7)$ and $b = \log_{14}(5)$, find $\log_{35}(28)$ in terms of $a$ and $b$.

One of the reasons I like this puzzle is that I did it a somewhat brutal way, and once I had the answer, a much neater way jumped out at me.

## The brutal way

Working whatever base I like, I have $a = \frac{\log(7)}{\log(7) + \log(2)}$ and $b = \frac{\log(5)}{\log(7) + \log(2)}$ - and I want to work out $\frac{\log(35)}{\log(28)}$, or $\frac{2\log(2) + \log(7)}{\log(5) + \log(7)}$.

I can see an awful lot of $\log(7)$, so how about working base 7?

Then $a = \frac{1}{1 + \log_7(2)}$, $b = \frac{\log_7{5}}{1 + \log_7(2)}$, and we want to know $\frac{2\log_7(2) + 1}{\log_7(5) + 1}$.

We can find $\log_7(2) = \frac{1}{a}-1$,

Substituting that in the second gives $\log_7(5) = \frac{b}{a}$.

So, the thing we need is $\frac{2\br{\frac{1}{a}-1} + 1}{\frac{b}{a} + 1}$.

Multiplying top and bottom by $a$ gives $\frac{2 - a}{ b + a }$.

## The d’oh! moment

Knowing the answer makes it much easier to find! Working base 14, we can say “we need $\log_{14}(28)$, which is $\log_{14}(196) - \log_{14}(7)$, or $2-a$.

Similarly, the bottom is $\log_{14}(5) + \log_{14}(7) = a+b$.

(Even simpler, as Mark points out: $\log_{14}(2) = 1 - \log_{14}(7)$. But we’ll let that slide.

### Woosh!

“$\ln(28) = \ln(35) + \ln(4) - \ln(5) \approx \ln(35) - 0.22$, so $\log{35}(28) \approx 1 - \frac{0.22}{\ln(35)}$. Meanwhile, $\ln(35)$ is close to $2\ln(6) \approx 3.6$. Now, $\frac{2.2}{36} \approx 0.06$, so we’re looking at $0.94$.”

“Is that the closest you can get it?”