Use the properties of logarithms to expand x x − 11 \frac { \sqrt { x } } { x - 11 } x − 11 x .A) 1 2 ln ⁡ x − ln ⁡ ( x − 11 ) \frac { 1 } { 2 } \ln x - \ln ( x - 11 ) 2 1 ln x − ln ( x − 11 ) B) ln ⁡ ( 1 2 x − x 11 ) \ln \left( \frac { 1 } { 2 } x - \frac { x } { 11 } \right) ln ( 2 1 x − 11 x ) C) ( ln ⁡ x ) 1 / 2 ln ⁡ x − ln ⁡ 11 \frac { ( \ln x ) ^ { 1 / 2 } } { \ln x - \ln 11 } l n x − l n 11 ( l n x ) 1/2 D) ( ln ⁡ x ) 1 / 2 − ln ⁡ ( x − 11 ) ( \ln x ) ^ { 1 / 2 } - \ln ( x - 11 ) ( ln x ) 1/2 − ln ( x − 11 ) E) 1 2 ln ⁡ x ⋅ ( ln ⁡ x − ln ⁡ 11 ) \frac { 1 } { 2 } \ln x \cdot ( \ln x - \ln 11 ) 2 1 ln x ⋅ ( ln x − ln 11 )

Question

Use the properties of logarithms to expand   x x − 11 \frac { \sqrt { x } } { x - 11 } x − 11 x ​ ​   .A)    1 2 ln ⁡ x − ln ⁡ ( x − 11 )  \frac { 1 } { 2 } \ln x - \ln ( x - 11 )  2 1 ​ ln x − ln ( x − 11 )  B)    ln ⁡ ( 1 2 x − x 11 )  \ln \left( \frac { 1 } { 2 } x - \frac { x } { 11 } \right)  ln ( 2 1 ​ x − 11 x ​ )  C)    ( ln ⁡ x )  1 / 2 ln ⁡ x − ln ⁡ 11 \frac { ( \ln x )  ^ { 1 / 2 } } { \ln x - \ln 11 } l n x − l n 11 ( l n x )  1/2 ​ D)    ( ln ⁡ x )  1 / 2 − ln ⁡ ( x − 11 )  ( \ln x )  ^ { 1 / 2 } - \ln ( x - 11 )  ( ln x )  1/2 − ln ( x − 11 )  E)    1 2 ln ⁡ x ⋅ ( ln ⁡ x − ln ⁡ 11 )  \frac { 1 } { 2 } \ln x \cdot ( \ln x - \ln 11 )  2 1 ​ ln x ⋅ ( ln x − ln 11 )

Sabah Al sabah · Accepted Answer

Final Answer : A, Explanation :   The correct expansion using properties of logarithms is   1 2 ln ⁡ x − ln ⁡ ( x − 11 ) \frac { 1 } { 2 } \ln x - \ln ( x - 11 ) 2 1 ​ ln x − ln ( x − 11 )  . This is because the square root can be expressed as a power of   1 / 2 1/2 1/2  , and the division inside the logarithm can be expressed as subtraction of logarithms.

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