Use the properties of logarithms to expand ln ⁡ 2 x x − 6 \ln \sqrt { \frac { 2 x } { x - 6 } } ln x − 6 2 x .A) ( ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 ) ) 1 / 2 ( \ln 2 + \ln x - \ln ( x - 6 ) ) ^ { 1 / 2 } ( ln 2 + ln x − ln ( x − 6 ) ) 1/2 B) ( ln ⁡ 2 ⋅ ln ⁡ x ln ⁡ x − ln ⁡ 6 ) 1 / 2 \left( \frac { \ln 2 \cdot \ln x } { \ln x - \ln 6 } \right) ^ { 1 / 2 } ( l n x − l n 6 l n 2 ⋅ l n x ) 1/2 C) 1 2 ( ln ⁡ 2 + ln ⁡ x − ln ⁡ x ln ⁡ 6 ) \frac { 1 } { 2 } \left( \ln 2 + \ln x - \frac { \ln x } { \ln 6 } \right) 2 1 ( ln 2 + ln x − l n 6 l n x ) D) 1 2 ( ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 ) ) \frac { 1 } { 2 } ( \ln 2 + \ln x - \ln ( x - 6 ) ) 2 1 ( ln 2 + ln x − ln ( x − 6 ) ) E) 1 2 ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 ) \frac { 1 } { 2 } \ln 2 + \ln x - \ln ( x - 6 ) 2 1 ln 2 + ln x − ln ( x − 6 )

Question

Use the properties of logarithms to expand   ln ⁡ 2 x x − 6 \ln \sqrt { \frac { 2 x } { x - 6 } } ln x − 6 2 x ​ ​   .A)    ( ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 )  )  1 / 2 ( \ln 2 + \ln x - \ln ( x - 6 )  )  ^ { 1 / 2 } ( ln 2 + ln x − ln ( x − 6 )  )  1/2 B)    ( ln ⁡ 2 ⋅ ln ⁡ x ln ⁡ x − ln ⁡ 6 )  1 / 2 \left( \frac { \ln 2 \cdot \ln x } { \ln x - \ln 6 } \right)  ^ { 1 / 2 } ( l n x − l n 6 l n 2 ⋅ l n x ​ )  1/2 C)    1 2 ( ln ⁡ 2 + ln ⁡ x − ln ⁡ x ln ⁡ 6 )  \frac { 1 } { 2 } \left( \ln 2 + \ln x - \frac { \ln x } { \ln 6 } \right)  2 1 ​ ( ln 2 + ln x − l n 6 l n x ​ )  D)    1 2 ( ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 )  )  \frac { 1 } { 2 } ( \ln 2 + \ln x - \ln ( x - 6 )  )  2 1 ​ ( ln 2 + ln x − ln ( x − 6 ) )  E)    1 2 ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 )  \frac { 1 } { 2 } \ln 2 + \ln x - \ln ( x - 6 )  2 1 ​ ln 2 + ln x − ln ( x − 6 )

Minh-Tu Nguyen · Accepted Answer

Final Answer : D, Explanation :   We can use the properties of logarithms to simplify the given expression: ln ⁡ 2 x x − 6 = 1 2 ln ⁡ ( 2 x x − 6 ) = 1 2 ( ln ⁡ 2 x − ln ⁡ ( x − 6 ) ) = 1 2 ln ⁡ 2 + 1 2 ln ⁡ x − 1 2 ln ⁡ ( x − 6 ) = 1 2 ( ln ⁡ 2 + ln ⁡ x − ln ⁡ ( x − 6 ) )  \ln \sqrt{\frac{2x}{x-6}}  & = \frac{1}{2} \ln \left(\frac{2x}{x-6}ight) \ & = \frac{1}{2} \left(\ln 2x - \ln (x-6)ight) \ & = \frac{1}{2} \ln 2 + \frac{1}{2} \ln x - \frac{1}{2} \ln (x-6) \ & = \frac{1}{2} (\ln 2 + \ln x - \ln(x-6))  ln x − 6 2 x ​ ​ ​ = 2 1 ​ ln ( x − 6 2 x ​ ) = 2 1 ​ ( ln 2 x − ln ( x − 6 ) ) = 2 1 ​ ln 2 + 2 1 ​ ln x − 2 1 ​ ln ( x − 6 ) = 2 1 ​ ( ln 2 + ln x − ln ( x − 6 )) ​ So, the correct answer is D.

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