Question

What does multiplication translate to when using logs?

Answer 1

General Idea:

We can use the product rule for exponentiation to derive a corresponding product rule for logarithms.

$e^a \cdot e^b=e^{a+b}$

Also we know$e^{ln(m)}=m$

Using the above rule we can write the below equation...

$e^{ln(xy)}=xy \; \rightarrow 1^{st} \; equation\\ \\ e^{ln(x)}=x \; \rightarrow \; 2^{nd} \; equation\\ \\ e^{ln(y)}=y \; \rightarrow \; 3^{rd} \; equation\\ \\ Substituting \; 2^{st} \; equation \; and \; 3^{rd} \; equation\; in\; 1^{st}\; equation, \; we\; get...\\ \\ e^{ln(xy)}=e^{ln(x)} \cdot e^{ln(y)}\\ \\ Applying \; the \; formula\; of\; exponent\; in\; right\; sides\; of\; equation\\ \\ e^{ln(xy)}=e^{ln(x)+ln(y)}\\ \\ Taking \; natural \; logarithm\; on\; both\; sides$

$ln[e^{ln(xy)}]=ln[e^{ln(x)+ln(y)}]\\ \\ And \; using\; the\; formula\; lnm^n=n(lnm), we\; get...\\ \\ ln(xy)=ln(x)+ln(y)$

Conclusion:

The product rule for exponentiation

$e^{x} \cdot e^{y}=e^{x+y}$

The product rule of logarithms

$ln(xy)=ln(x)+ln(y)$

Answer 2

It translates to addition.