Logarithm - Basics

log of any number is defined for positive real numbers only,so |b| and b are same. so that answer is lna

Using properties of logarithms, we have

$\ln ab - \ln |b| = \ln \dfrac{ab}{|b|}$

Since for the original expression to be defined over a certain set of real numbers, we need both $\ln ab$ and $\ln |b|$ to be defined. $\ln |b|$ is always defined for $b \neq 0$ while $\ln ab$ is defined only when $ab > 0$ , i.e, both positive or both negative.

• If $b>0$ then $\ln \dfrac{ab}{|b|} = \ln a$ which is defined since $a>0$ if $b>0$ .

• If $b<0$ then $\ln \dfrac{ab}{|b|} = \ln (-a)$ which is defined since $a<0$ if $b<0$ .

The function that corresponds to $f(a) := \begin{cases} a &, a>0 \\ -a &, a<0 \end{cases} \$ is the function $f(a) = |a|$ . Hence the required answer is $\boxed{\ln |a|}$ .

log (x) with base 'e' is defined only when x is greater than zero. now consider the case when b>0, then 'a' must also be greater than zero ,since log (a b) is defined only for a b>0. so given expression becomes log(a*b/b)= log (a). if 'b' <0, then 'a' must also be negative. hence ans should be log (mod a).

log e (a)= ln (a),, ln(ab) can be either ln(a)+ln(b) or ln(-a)+ln(-b), to satisfy both cases we have to choose loge|a|

ln ab= ln a + ln b ln a/b = ln a - ln b

' ln '(log with base e) is only defined for natural numbers (positive numbers) and here we have ln |b| so it means b can take a negative value hence we got a modulus with it, but when its multiplied with a then it we don't have the modulus which infers that a and b have the same sign. therefore when we subtract ln |b| from ln ab then we are left with ln a but since it can also be negative if b is negative which will render our answer undefined so we put a modulus with it