Question:
The Napier logarithm can be defined as follows:
$$Nap.log~x = 10^7 \ln \frac{10^7}{x}$$
Prove the following rules:
a) $$Nap.log~xy = Nap.log~x + Nap.log~y - Nap.log~1$$
b) $$Nap.log~\frac{x}{y} = Nap.log~x - Nap.log~y + Nap.log~1$$
c) $$Nap.log~x^a = a \cdot Nap.log~x + (1-a) Nap.log~1$$
Attempted answer:
The basic approach is to use the definition of Napier logarithm to convert to natural logarithm, rearrange them and show that the results can be converted back to Napier logarithm to fulfill the rules to be proven.
a)
$$Nap.log~xy = 10^7 \ln \frac{10^7}{xy} = 10^7 \ln \frac{10^7}{x} + 10^7 \ln \frac{10^7}{y} - 10^7 \ln 10^7$$ $$ = Nap.log~x + Nap.log~y - Nap.log~1$$
b) For this law, I will just use the trick that x divided by y is the same as multiplying x with the reciprocal of y:
$$Nap.log~\frac{x}{y} = Nap.log~(x \cdot \frac{1}{y}) = Nap.log~x + Nap.log~\frac{1}{y} =$$ $$ Nap.log~x - Nap.log~y + Nap.log~1$$
c) This one is a bit tricky for me. I can tell it must involve using the standard logarithm for exponent and moving it down as a factor, but not entirely clear on how to apply it. Here is what I have so far:
$$Nap.log~a^x = 10^7 ln \frac{10^7}{x^a} = 10^7 \ln 10^7 - 10^7 \ln x^a = 10^7 \ln 10^7 - a \cdot 10^7 \ln x $$
...but this does not seem to easily work out. How do I wrap up this last part of the question?
For c), adding $$10^7a\ln 10^7-10^7a\ln 10^7\ (=0)$$ works.
$$\begin{align}Nap.log~x^a&=10^7\ln\frac{10^7}{x^a} \\\\&=10^7\ln 10^7-10^7a\ln x \\\\&=10^7\ln 10^7-10^7a\ln x+10^7a\ln 10^7-10^7a\ln 10^7 \\\\&=(10^7a\ln 10^7-10^7a\ln x)+(10^7\ln 10^7-10^7a\ln 10^7) \\\\&=10^7a\ln\frac{10^7}{x}+(1-a)10^7\ln 10^7 \\\\&=a \cdot Nap.log~x + (1-a) Nap.log~1\end{align}$$