log-ab-log-b-A-log-a-B-log-a-C-log-a-D-None-of-these-

Question Number 113800 by Aina Samuel Temidayo last updated on 15/Sep/20

\log (ab) - \log ∣ b ∣ =

A . \log (a) B . \log ∣ a ∣ C . - \log (a) D .

None of these .

Answered by MJS_new last updated on 15/Sep/20

(1) a < 0 \land b < 0

\log a b - \log ∣ b ∣ = \log (- a)

(2) a > 0 \land b > 0

\log a b - \log ∣ b ∣ = \log a

\Rightarrow answer is B . \log ∣ a ∣

Commented by MJS_new last updated on 15/Sep/20

test it .

i . e . a = - 3 \land b = - 5

\log a b = \log 15

\log ∣ b ∣ = \log 5

\log 15 - \log 5 = \log 3 = \log - a

\Rightarrow if a < 0 \land b < 0 we get \log - a (but a < 0 \Rightarrow - a > 0)

if a > 0 \land b > 0 we get \log a

\Rightarrow we get \log ∣ a ∣

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

\log (ab) = \log (a) + \log (b)

\Rightarrow a, b is + ve

\Rightarrow \log ∣ b ∣ = \log b

\Rightarrow \log (ab) - \log ∣ b ∣

= \log (ab) - \log (b)

= \log (\frac{ab}{b}) = \log (a) .

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

a, b cannot be - ve .

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

From (1), you said a < 0 \land b < 0

\Rightarrow ab > 0

Note that ∣ b ∣ > 0

\Rightarrow \frac{ab}{∣ b ∣} is + ve

Commented by MJS_new last updated on 15/Sep/20

you misunderstand ∣ x ∣ it seems . remember

∣ x ∣= {\begin{cases} - x; x < 0 \\ x; x ⩾ 0 \end{cases}

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

since \log (ab) = \log (a) + \log (b)

a, b should not be negative . .

Commented by MJS_new last updated on 15/Sep/20

no . \log a b = \log a + \log b only if a > 0 \land b > 0

\log a b exists if a b > 0 \Rightarrow a > 0 \land b > 0 \lor a < 0 \land b < 0

if a < 0 \land b < 0 \Rightarrow \log a b \neq \log a + \log b

but we can still calculate \log a b - \log ∣ b ∣

because a b > 0 \land ∣ b ∣> 0 and we get \log ∣ a ∣

because a < 0 but the argument we get is > 0

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

Ok . Thanks \overset{}{.}