Question Number 193230 by MATHEMATICSAM last updated on 07/Jun/23
$$\boldsymbol{\mathrm{Choose}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{option}}: \\ $$$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{consecutive}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{and}\:\mathrm{log}\left(\mathrm{1}\:+\:{ac}\right)\:=\:\mathrm{2}{k}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is}: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{log}\:{a} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{log}\:{b} \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2} \\ $$$$\left.\mathrm{d}\right)\:\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Give}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{explaination}}\:\boldsymbol{\mathrm{also}}. \\ $$
Answered by Rajpurohith last updated on 08/Jun/23
$${let}\:{a}={b}−\mathrm{1}\:;\:{c}={b}+\mathrm{1}\:\left({since}\:{a}\:,{b}\:{and}\:{c}\:{are}\:{consecutive}.\right) \\ $$$${so}\:\mathrm{1}+{ac}=\mathrm{1}+\left({b}−\mathrm{1}\right)\left({b}+\mathrm{1}\right)=\mathrm{1}+{b}^{\mathrm{2}} −\mathrm{1}={b}^{\mathrm{2}} \\ $$$$\Rightarrow{log}\left(\mathrm{1}+{ac}\right)={log}\left({b}^{\mathrm{2}} \right)=\mathrm{2}{log}\left({b}\right)=\mathrm{2}{k} \\ $$$${hence}\:{k}={log}\left({b}\right). \\ $$