Ask Uncle Colin: How did they get ln⁡(50)?

Dear Uncle Colin,

I get $-\frac{\ln (0.02)}{0.03}$ as my answer to a question. They have $\frac{100\ln (50)}{3}$. Numerically, they seem to be the same, but they look completely different. What gives?

-- Polishing Off Weird Exponents, Really Stuck

Dear POWERS,

What you need here are the log laws (to show that $-\ln (0.02)=\ln (50)$, and a bit of fractions work (to show that $\frac{1}{0.03}=\frac{100}{3}$.

Let’s do the second bit first; multiply the top and bottom of $\frac{1}{0.03}$ by 100 and you get $\frac{100}{3}$. Simple.

As for $-\ln (0.02)$, there are several ways to do it. One is to think of it as $(-1)\ln (0.02)$, which is the same as $\ln ((0.02{)}^{-1})=\ln \left({\left(\frac{2}{100}\right)}^{-1}\right)=\ln \left(50\right)$.

Another is to write it as a fraction immediately: $-\ln (0.02)=-\ln \left(\frac{2}{100}\right)=-[\ln (2)-\ln (100)]=\ln (100)-\ln (2)=\ln (50)$.

It takes a bit of practise to get used to seeing what decimals can be written more neatly in another form – personally, I recommend using fractions ahead of decimals pretty much always.

-- Uncle Colin