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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.
$$\mathrm{log}\:\left(\mathrm{ab}\right)−\mathrm{log}\mid\mathrm{b}\mid\:= \\ $$$$ \\ $$$$\mathrm{A}.\:\mathrm{log}\left(\mathrm{a}\right)\:\mathrm{B}.\:\mathrm{log}\:\mid\mathrm{a}\mid\:\mathrm{C}.\:−\mathrm{log}\left(\mathrm{a}\right)\:\mathrm{D}. \\ $$$$\mathrm{None}\:\mathrm{of}\:\mathrm{these}. \\ $$
Answered by MJS_new last updated on 15/Sep/20
(1) a<0∧b<0 log ab −log ∣b∣ =log (−a) (2) a>0∧b>0 log ab −log ∣b∣ =log a ⇒ answer is B. log ∣a∣
$$\left(\mathrm{1}\right)\:{a}<\mathrm{0}\wedge{b}<\mathrm{0} \\ $$$$\mathrm{log}\:{ab}\:−\mathrm{log}\:\mid{b}\mid\:=\mathrm{log}\:\left(−{a}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}>\mathrm{0}\wedge{b}>\mathrm{0} \\ $$$$\mathrm{log}\:{ab}\:−\mathrm{log}\:\mid{b}\mid\:=\mathrm{log}\:{a} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:{B}.\:\mathrm{log}\:\mid{a}\mid \\ $$
Commented by MJS_new last updated on 15/Sep/20
test it. i.e. a=−3∧b=−5 log ab =log 15 log ∣b∣ =log 5 log 15 −log 5 =log 3 =log −a ⇒ if a<0∧b<0 we get log −a (but a<0 ⇒ −a>0) if a>0∧b>0 we get log a ⇒ we get log ∣a∣
$$\mathrm{test}\:\mathrm{it}. \\ $$$$\mathrm{i}.\mathrm{e}.\:{a}=−\mathrm{3}\wedge{b}=−\mathrm{5} \\ $$$$\mathrm{log}\:{ab}\:=\mathrm{log}\:\mathrm{15} \\ $$$$\mathrm{log}\:\mid{b}\mid\:=\mathrm{log}\:\mathrm{5} \\ $$$$\mathrm{log}\:\mathrm{15}\:−\mathrm{log}\:\mathrm{5}\:=\mathrm{log}\:\mathrm{3}\:=\mathrm{log}\:−{a} \\ $$$$\Rightarrow\:\mathrm{if}\:{a}<\mathrm{0}\wedge{b}<\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{log}\:−{a}\:\left(\mathrm{but}\:{a}<\mathrm{0}\:\Rightarrow\:−{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\mathrm{if}\:{a}>\mathrm{0}\wedge{b}>\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{log}\:{a} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{get}\:\mathrm{log}\:\mid{a}\mid \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
log(ab) = log(a)+log(b) ⇒a,b is +ve ⇒ log∣b∣ = log b ⇒ log(ab) −log∣b∣ =log(ab)−log (b) =log(((ab)/b)) = log(a).
$$\mathrm{log}\left(\mathrm{ab}\right)\:=\:\mathrm{log}\left(\mathrm{a}\right)+\mathrm{log}\left(\mathrm{b}\right)\: \\ $$$$\Rightarrow\mathrm{a},\mathrm{b}\:\mathrm{is}\:+\mathrm{ve} \\ $$$$\Rightarrow\:\mathrm{log}\mid\mathrm{b}\mid\:=\:\mathrm{log}\:\mathrm{b}\: \\ $$$$\Rightarrow\:\mathrm{log}\left(\mathrm{ab}\right)\:−\mathrm{log}\mid\mathrm{b}\mid\: \\ $$$$=\mathrm{log}\left(\mathrm{ab}\right)−\mathrm{log}\:\left(\mathrm{b}\right) \\ $$$$=\mathrm{log}\left(\frac{\mathrm{ab}}{\mathrm{b}}\right)\:=\:\mathrm{log}\left(\mathrm{a}\right). \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
a,b cannot be −ve.
$$\mathrm{a},\mathrm{b}\:\mathrm{cannot}\:\mathrm{be}\:−\mathrm{ve}. \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
From (1), you said a<0∧b<0 ⇒ ab >0 Note that ∣b∣ >0 ⇒ ((ab)/(∣b∣)) is +ve
$$\mathrm{From}\:\left(\mathrm{1}\right),\:\mathrm{you}\:\mathrm{said}\:\mathrm{a}<\mathrm{0}\wedge\mathrm{b}<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{ab}\:>\mathrm{0} \\ $$$$\mathrm{Note}\:\mathrm{that}\:\mid\mathrm{b}\mid\:>\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{ab}}{\mid\mathrm{b}\mid}\:\mathrm{is}\:+\mathrm{ve} \\ $$$$ \\ $$
Commented by MJS_new last updated on 15/Sep/20
you misunderstand ∣x∣ it seems. remember ∣x∣= { ((−x; x<0)),((x; x≥0)) :}
$$\mathrm{you}\:\mathrm{misunderstand}\:\mid{x}\mid\:\mathrm{it}\:\mathrm{seems}.\:\mathrm{remember} \\ $$$$\mid{x}\mid=\begin{cases}{−{x};\:{x}<\mathrm{0}}\\{{x};\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
since log(ab) = log(a)+log(b) a,b should not be negative..
$$\mathrm{since}\:\mathrm{log}\left(\mathrm{ab}\right)\:=\:\mathrm{log}\left(\mathrm{a}\right)+\mathrm{log}\left(\mathrm{b}\right) \\ $$$$\mathrm{a},\mathrm{b}\:\mathrm{should}\:\mathrm{not}\:\mathrm{be}\:\mathrm{negative}.. \\ $$
Commented by MJS_new last updated on 15/Sep/20
no. log ab =log a +log b only if a>0∧b>0 log ab exists if ab>0 ⇒ a>0∧b>0 ∨ a<0∧b<0 if a<0∧b<0 ⇒ log ab ≠ log a +log b but we can still calculate log ab −log ∣b∣ because ab>0∧∣b∣>0 and we get log ∣a∣ because a<0 but the argument we get is >0
$$\mathrm{no}.\:\mathrm{log}\:{ab}\:=\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:\mathrm{only}\:\mathrm{if}\:{a}>\mathrm{0}\wedge{b}>\mathrm{0} \\ $$$$\mathrm{log}\:{ab}\:\mathrm{exists}\:\mathrm{if}\:{ab}>\mathrm{0}\:\Rightarrow\:{a}>\mathrm{0}\wedge{b}>\mathrm{0}\:\vee\:{a}<\mathrm{0}\wedge{b}<\mathrm{0} \\ $$$$\mathrm{if}\:{a}<\mathrm{0}\wedge{b}<\mathrm{0}\:\Rightarrow\:\mathrm{log}\:{ab}\:\neq\:\mathrm{log}\:{a}\:+\mathrm{log}\:{b} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{still}\:\mathrm{calculate}\:\mathrm{log}\:{ab}\:−\mathrm{log}\:\mid{b}\mid \\ $$$$\mathrm{because}\:{ab}>\mathrm{0}\wedge\mid{b}\mid>\mathrm{0}\:\mathrm{and}\:\mathrm{we}\:\mathrm{get}\:\mathrm{log}\:\mid{a}\mid \\ $$$$\mathrm{because}\:{a}<\mathrm{0}\:\mathrm{but}\:\mathrm{the}\:\mathrm{argument}\:\mathrm{we}\:\mathrm{get}\:\mathrm{is}\:>\mathrm{0} \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
Ok. Thanks.^
$$\mathrm{Ok}.\:\mathrm{Thanks}\overset{} {.} \\ $$

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