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If-a-log-24-12-b-log-36-24-and-c-log-48-36-then-1-abc-is-equal-to-A-2ab-B-2ac-C-2bc-D-0-




Question Number 113803 by Aina Samuel Temidayo last updated on 15/Sep/20
If a=log_(24) 12, b=log_(36) 24 and  c=log_(48) 36, then 1+abc is equal to    (A) 2ab (B) 2ac (C) 2bc (D) 0
$$\mathrm{If}\:\mathrm{a}=\mathrm{log}_{\mathrm{24}} \mathrm{12},\:\mathrm{b}=\mathrm{log}_{\mathrm{36}} \mathrm{24}\:\mathrm{and} \\ $$$$\mathrm{c}=\mathrm{log}_{\mathrm{48}} \mathrm{36},\:\mathrm{then}\:\mathrm{1}+\mathrm{abc}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:\mathrm{2ab}\:\left(\mathrm{B}\right)\:\mathrm{2ac}\:\left(\mathrm{C}\right)\:\mathrm{2bc}\:\left(\mathrm{D}\right)\:\mathrm{0} \\ $$
Answered by bemath last updated on 15/Sep/20
a = ((ln 12)/(ln 24))=((ln 12)/(ln 2+ln 12)) , b=((ln 24)/(ln 36))=((ln 2+ln 12)/(ln 3+ln 12))  c=((ln 3+ln 12)/(ln 4+ln 12))  a×b×c = ((ln 12)/(ln 2+ln 12))×((ln 2+ln 12)/(ln 3+ln 12))×((ln 3+ln 12)/(ln 4+ln 12))              = ((ln 12)/(ln 4+ln 12)) = log _(48) (12)  1+abc = 1+log _(48) (12)                = log _(48) (12×48)                = ((ln 4+2.ln 12)/(ln 4+ln 12))          = ((2ln 2+2ln 12)/(ln 4+ln 12)) = 2(((ln 2+ln 12)/(ln 4+ln 12)))          = 2bc
$${a}\:=\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{24}}=\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{12}}\:,\:{b}=\frac{\mathrm{ln}\:\mathrm{24}}{\mathrm{ln}\:\mathrm{36}}=\frac{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{12}} \\ $$$${c}=\frac{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}} \\ $$$${a}×{b}×{c}\:=\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{12}}×\frac{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{12}}×\frac{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}}\:=\:\mathrm{log}\:_{\mathrm{48}} \left(\mathrm{12}\right) \\ $$$$\mathrm{1}+{abc}\:=\:\mathrm{1}+\mathrm{log}\:_{\mathrm{48}} \left(\mathrm{12}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{log}\:_{\mathrm{48}} \left(\mathrm{12}×\mathrm{48}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{ln}\:\mathrm{4}+\mathrm{2}.\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2ln}\:\mathrm{2}+\mathrm{2ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}}\:=\:\mathrm{2}\left(\frac{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{12}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{2}{bc} \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$
Answered by som(math1967) last updated on 15/Sep/20
C)2bc
$$\left.\mathrm{C}\right)\mathrm{2bc} \\ $$
Answered by $@y@m last updated on 15/Sep/20
24^a =12 .....(A)  36^b =24 .....(B)  48^c =36 .....(C)  From (C),  (48^c )^b =36^b   48^(bc) =24  {usimg (B)  (48^(bc) )^a =24^a   48^(abc) =12  {using (A)  48^(abc) ×48=12×48  48^(1+abc) =24^2   48^(1+abc) =36^(2b)   48^(1+abc) =48^(2bc)   1+abc=2bc
$$\mathrm{24}^{{a}} =\mathrm{12}\:…..\left({A}\right) \\ $$$$\mathrm{36}^{{b}} =\mathrm{24}\:…..\left({B}\right) \\ $$$$\mathrm{48}^{{c}} =\mathrm{36}\:…..\left({C}\right) \\ $$$${From}\:\left({C}\right), \\ $$$$\left(\mathrm{48}^{{c}} \right)^{{b}} =\mathrm{36}^{{b}} \\ $$$$\mathrm{48}^{{bc}} =\mathrm{24}\:\:\left\{{usimg}\:\left({B}\right)\right. \\ $$$$\left(\mathrm{48}^{{bc}} \right)^{{a}} =\mathrm{24}^{{a}} \\ $$$$\mathrm{48}^{{abc}} =\mathrm{12}\:\:\left\{{using}\:\left({A}\right)\right. \\ $$$$\mathrm{48}^{{abc}} ×\mathrm{48}=\mathrm{12}×\mathrm{48} \\ $$$$\mathrm{48}^{\mathrm{1}+{abc}} =\mathrm{24}^{\mathrm{2}} \\ $$$$\mathrm{48}^{\mathrm{1}+{abc}} =\mathrm{36}^{\mathrm{2}{b}} \\ $$$$\mathrm{48}^{\mathrm{1}+{abc}} =\mathrm{48}^{\mathrm{2}{bc}} \\ $$$$\mathrm{1}+{abc}=\mathrm{2}{bc} \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 15/Sep/20
Nice method. Thanks.
$$\mathrm{Nice}\:\mathrm{method}.\:\mathrm{Thanks}. \\ $$

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