Question

Let p be an odd prime and r a primitive root of p. Show that if a and b are positive integers in Zp,the${\log }_r\left(ab\right)\equiv {\log }_r\left(a\right)+{\log }_{r}b(modp-1)$

Short answer

${\log }_r\left(ab\right)\equiv {\log }_r\left(a\right)+{\log }_{r}b(modp-1)$

Step-by-step solution

Step: 1

$\begin{array}{r}{\log }_r\left(ab\right)=x\\ \Rightarrow r^{x}modp=ab\\ {\log }_r\left(a\right)=y\\ \Rightarrow r^{y}modp=a\\ {\log }_r\left(b\right)=z\\ \Rightarrow r^{z}modp=b\end{array}$

Step: 2

We obtain

$\begin{array}{r}ab=\left(r^{y}modp\right)\left(r^{2}modp\right)\\ =\left(r^y{r}^z\right)modp\\ =r^{y+z}modp\end{array}$

Step: 3

$$\begin{gathered} \sin ce{r}^{x}modp=ab \\ \mathrm{Hence}\mathrm{we}\mathrm{get} \\ {\log }_{\mathrm{r}}\left(\mathrm{ab}\right)\equiv {\log }_{\mathrm{r}}\left(\mathrm{a}\right)+{\log }_{\mathrm{r}}\mathrm{b}(mod\mathrm{p}-1) \end{gathered}$$