Question Number 65971 by gunawan last updated on 07/Aug/19
![If (a^2 /b^2 )=12 then log(^3 (√(b/a)))=.. a. −2 b. −1 c. 0 d. 1 e. 2](https://www.tinkutara.com/question/Q65971.png)
$$\mathrm{If}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{12}\:\mathrm{then}\:\mathrm{log}\left(\:^{\mathrm{3}} \sqrt{\frac{{b}}{{a}}}\right)=.. \\ $$$${a}.\:−\mathrm{2} \\ $$$${b}.\:−\mathrm{1} \\ $$$${c}.\:\mathrm{0} \\ $$$${d}.\:\mathrm{1} \\ $$$${e}.\:\mathrm{2} \\ $$
Answered by MJS last updated on 07/Aug/19
![ln ((√(12)))^(1/3) =ln ((12))^(1/6) =((ln 12)/6)=((ln 2)/3)+((ln 3)/6)](https://www.tinkutara.com/question/Q65976.png)
$$\mathrm{ln}\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{12}}}\:=\mathrm{ln}\:\sqrt[{\mathrm{6}}]{\mathrm{12}}\:=\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{6}}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{6}} \\ $$
Answered by kaivan.ahmadi last updated on 07/Aug/19
![(a^2 /b^2 )=12⇒((b/a))^2 =(1/(12))⇒(b/a)=(√(1/(12)))⇒ ((b/a))^(1/3) =((1/(12)))^(1/6) =((1/(12)))^(1/6) =12^(−(1/6)) ⇒ log(((b/a))^(1/3) )=log12^(−(1/6)) =−(1/6)log12=−log_6 12 =−(log_6 6+log_6 2)=−1−log_6 2](https://www.tinkutara.com/question/Q66012.png)
$$\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{12}\Rightarrow\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{12}}\Rightarrow\frac{{b}}{{a}}=\sqrt{\frac{\mathrm{1}}{\mathrm{12}}}\Rightarrow \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{b}}{{a}}}=\sqrt[{\mathrm{6}}]{\frac{\mathrm{1}}{\mathrm{12}}}=\left(\frac{\mathrm{1}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{12}^{−\frac{\mathrm{1}}{\mathrm{6}}} \Rightarrow \\ $$$${log}\left(\sqrt[{\mathrm{3}}]{\frac{{b}}{{a}}}\right)={log}\mathrm{12}^{−\frac{\mathrm{1}}{\mathrm{6}}} =−\frac{\mathrm{1}}{\mathrm{6}}{log}\mathrm{12}=−{log}_{\mathrm{6}} \mathrm{12} \\ $$$$=−\left({log}_{\mathrm{6}} \mathrm{6}+{log}_{\mathrm{6}} \mathrm{2}\right)=−\mathrm{1}−{log}_{\mathrm{6}} \mathrm{2} \\ $$$$ \\ $$