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


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 $\ln\left( (0.02)^{-1} \right) = \ln\left( \left(\frac {2}{100}\right)^{-1} \right) = \ln \left( 50 \right)$.

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

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