General Idea:
We can use the product rule for exponentiation to derive a corresponding product rule for logarithms.
[tex] e^a \cdot e^b=e^{a+b} [/tex]
Also we know[tex] e^{ln(m)}=m [/tex]
Using the above rule we can write the below equation...
[tex] e^{ln(xy)}=xy \; \rightarrow 1^{st} \; equation\\ \\ e^{ln(x)}=x \; \rightarrow \; 2^{nd} \; equation\\ \\ e^{ln(y)}=y \; \rightarrow \; 3^{rd} \; equation\\ \\ Substituting \; 2^{st} \; equation \; and \; 3^{rd} \; equation\; in\; 1^{st}\; equation, \; we\; get...\\ \\ e^{ln(xy)}=e^{ln(x)} \cdot e^{ln(y)}\\ \\ Applying \; the \; formula\; of\; exponent\; in\; right\; sides\; of\; equation\\ \\ e^{ln(xy)}=e^{ln(x)+ln(y)}\\ \\ Taking \; natural \; logarithm\; on\; both\; sides\\ \\ [/tex]
[tex] ln[e^{ln(xy)}]=ln[e^{ln(x)+ln(y)}]\\ \\ And \; using\; the\; formula\; lnm^n=n(lnm), we\; get...\\ \\ ln(xy)=ln(x)+ln(y) [/tex]
Conclusion:
The product rule for exponentiation
[tex] e^{x} \cdot e^{y}=e^{x+y} [/tex]
The product rule of logarithms
[tex] ln(xy)=ln(x)+ln(y) [/tex]
